kode python

Bagaimana Anda membalikkan angka ke belakang dengan python?

Program python berikut membalikkan angka multi-digit yang diberikan. Dalam contoh ini, kami hanya mencetak angka dalam urutan terbalik tanpa menghilangkan nol berlebihan. Misalnya, jika angka yang dimasukkan adalah "1000", angka terbalik yang ditampilkan adalah "0001", bukan "1"

number = int(input("Enter a number: ")) reverse_number = '' while number > 0: last_digit = number % 10 reverse_number = reverse_number + str(last_digit) number = number // 10 print("Reverse number is: ", reverse_number)

Program di atas menggunakan logika berikut untuk membalikkan angka,

Ekstrak digit terakhir menggunakan operator modulus (Sisa angka setelah dibagi 10)
Tambahkan digit terakhir sebagai digit pertama dalam string terbalik
Sekarang hapus digit terakhir dari angka asli yang membaginya dengan 10 (perhatikan penggunaan operator pembagian bilangan bulat //. Operator / jika digunakan akan menghasilkan pecahan. )
Ulangi dari langkah 1 hingga nomor asli tidak memiliki digit yang tersisa

Program berikut menggunakan pendekatan yang sedikit berbeda. Dalam contoh ini, kami menyimpan angka terbalik sebagai bilangan bulat, bukan string. Hal ini memastikan bahwa saat kita membalikkan "1000", kita mendapatkan "1", bukan "0001". Perhatikan bahwa sebelum menjumlahkan digit terakhir, kita mengalikan angka sebelumnya dengan 10 untuk menaikkan nilai tempatnya

Reverse of no. is 26542Reverse of no. is 12345607Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num85Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num64 Reverse of no. is 12345639Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num77 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num24Reverse of no. is 26541_______58Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num83 ________2______10

Reverse of no. is 26542Reverse of no. is 12345610Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num83Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num64 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num24

Reverse of no. is 123456_53

Reverse of no. is 12345654Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num64 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num51

Reverse of no. is 12345661Reverse of no. is 12345662

Reverse of no. is 123456_63

C#

Reverse of no. is 123456_64

654321_37

Reverse of no. is 2654_05

Reverse of no. is 1234562 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num99

#include <stdio.h>8 #include <stdio.h>9

Reverse of no. is 26542Reverse of no. is 265452 Reverse of no. is 26541

Reverse of no. is 26542Reverse of no. is 26547

Reverse of no. is 26542Reverse of no. is 26549 Reverse of no. is 26540

Reverse of no. is 26542Reverse of no. is 265416

Reverse of no. is 2654_7

#include <stdio.h>_5

Reverse of no. is 2654_21

PHP

Reverse of no. is 2654_52

Reverse of no. is 2654_23

Reverse of no. is 2654_61 Reverse of no. is 265462

Reverse of no. is 265426 Reverse of no. is 265427

Reverse of no. is 2654_28

Reverse of no. is 2654_29

Reverse of no. is 2654_55 Reverse of no. is 265439Reverse of no. is 265457Reverse of no. is 265458

Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num6Reverse of no. is 12345618 Reverse of no. is 265461Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num16

Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num6Reverse of no. is 12345618 Reverse of no. is 265426Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num16

Reverse of no. is 26542Reverse of no. is 265451Reverse of no. is 265457 Reverse of no. is 265453

Reverse of no. is 26542Reverse of no. is 265461 Reverse of no. is 265456Reverse of no. is 265457 Reverse of no. is 265458

Reverse of no. is 2654_59Reverse of no. is 265426Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num16

Reverse of no. is 26542Reverse of no. is 265426 Reverse of no. is 265464

Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num6Reverse of no. is 26549 Reverse of no. is 265461Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num16

Reverse of no. is 2654_7

Reverse of no. is 2654_90

Reverse of no. is 2654_57 Reverse of no. is 265492

Reverse of no. is 2654_93 Original no: 1234 Reversed no: 43216Reverse of no. is 265495

Reverse of no. is 2654_96Reverse of no. is 265439Reverse of no. is 265457Reverse of no. is 265499

Reverse of no. is 2654_82

Reverse of no. is 2654_01

Javascript

Reverse of no. is 2654_02

Reverse of no. is 2654_85

Reverse of no. is 2654_37

Reverse of no. is 2654_87 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num8

Reverse of no. is 2654_87 Reverse of no. is 265450

Reverse of no. is 2654_55 Reverse of no. is 265492

Reverse of no. is 2654_2Input: n (1) Intilize rev1=0, rev2=0 (2) Compute no of digits in given input n and store it in size variable. (3) Divide the number n into two parts i.e first_half and second_half. (4) Loop for i in range(mid): (a) Remainder of first_half by 10 add it to the multiplication of 10 and rev1. rem=first_half%10 rev1=10*rev1+rem (b) Remainder of second_half by 10 add it to the multiplication of 10 and rev2. rem=second_half%10 rev2=10*rev2+rem (c) Divide first_half and second_half with 10. first_half=first_half//10 second_half=second_half//10 (5)if size is even. return rev2*10**mid+rev1 otherwise return (rev2*10**mid+rev1)*10+first_half00

Reverse of no. is 26542Reverse of no. is 265459

Reverse of no. is 2654_7

Reverse of no. is 2654_90

Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num6Input: n (1) Intilize rev1=0, rev2=0 (2) Compute no of digits in given input n and store it in size variable. (3) Divide the number n into two parts i.e first_half and second_half. (4) Loop for i in range(mid): (a) Remainder of first_half by 10 add it to the multiplication of 10 and rev1. rem=first_half%10 rev1=10*rev1+rem (b) Remainder of second_half by 10 add it to the multiplication of 10 and rev2. rem=second_half%10 rev2=10*rev2+rem (c) Divide first_half and second_half with 10. first_half=first_half//10 second_half=second_half//10 (5)if size is even. return rev2*10**mid+rev1 otherwise return (rev2*10**mid+rev1)*10+first_half15Original no: 1234 Reversed no: 43216

#include <stdio.h>_5

Reverse of no. is 2654_34

KeluaranReverse of no. is 2654_

Kompleksitas Waktu. O(log(n))
Ruang Bantu. O(log(n)), di mana n adalah nomor masukan

Menggunakan Rekursi, tanpa variabel tambahan

C++

Reverse of no. is 1234562 Reverse of no. is 1234563 Reverse of no. is 1234564

Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num1 Input: n (1) Intilize rev1=0, rev2=0 (2) Compute no of digits in given input n and store it in size variable. (3) Divide the number n into two parts i.e first_half and second_half. (4) Loop for i in range(mid): (a) Remainder of first_half by 10 add it to the multiplication of 10 and rev1. rem=first_half%10 rev1=10*rev1+rem (b) Remainder of second_half by 10 add it to the multiplication of 10 and rev2. rem=second_half%10 rev2=10*rev2+rem (c) Divide first_half and second_half with 10. first_half=first_half//10 second_half=second_half//10 (5)if size is even. return rev2*10**mid+rev1 otherwise return (rev2*10**mid+rev1)*10+first_half29Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num1 Input: n (1) Intilize rev1=0, rev2=0 (2) Compute no of digits in given input n and store it in size variable. (3) Divide the number n into two parts i.e first_half and second_half. (4) Loop for i in range(mid): (a) Remainder of first_half by 10 add it to the multiplication of 10 and rev1. rem=first_half%10 rev1=10*rev1+rem (b) Remainder of second_half by 10 add it to the multiplication of 10 and rev2. rem=second_half%10 rev2=10*rev2+rem (c) Divide first_half and second_half with 10. first_half=first_half//10 second_half=second_half//10 (5)if size is even. return rev2*10**mid+rev1 otherwise return (rev2*10**mid+rev1)*10+first_half31

Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num6Input: n (1) Intilize rev1=0, rev2=0 (2) Compute no of digits in given input n and store it in size variable. (3) Divide the number n into two parts i.e first_half and second_half. (4) Loop for i in range(mid): (a) Remainder of first_half by 10 add it to the multiplication of 10 and rev1. rem=first_half%10 rev1=10*rev1+rem (b) Remainder of second_half by 10 add it to the multiplication of 10 and rev2. rem=second_half%10 rev2=10*rev2+rem (c) Divide first_half and second_half with 10. first_half=first_half//10 second_half=second_half//10 (5)if size is even. return rev2*10**mid+rev1 otherwise return (rev2*10**mid+rev1)*10+first_half34

Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num6Reverse of no. is 26549 Input: n (1) Intilize rev1=0, rev2=0 (2) Compute no of digits in given input n and store it in size variable. (3) Divide the number n into two parts i.e first_half and second_half. (4) Loop for i in range(mid): (a) Remainder of first_half by 10 add it to the multiplication of 10 and rev1. rem=first_half%10 rev1=10*rev1+rem (b) Remainder of second_half by 10 add it to the multiplication of 10 and rev2. rem=second_half%10 rev2=10*rev2+rem (c) Divide first_half and second_half with 10. first_half=first_half//10 second_half=second_half//10 (5)if size is even. return rev2*10**mid+rev1 otherwise return (rev2*10**mid+rev1)*10+first_half37Input: n (1) Intilize rev1=0, rev2=0 (2) Compute no of digits in given input n and store it in size variable. (3) Divide the number n into two parts i.e first_half and second_half. (4) Loop for i in range(mid): (a) Remainder of first_half by 10 add it to the multiplication of 10 and rev1. rem=first_half%10 rev1=10*rev1+rem (b) Remainder of second_half by 10 add it to the multiplication of 10 and rev2. rem=second_half%10 rev2=10*rev2+rem (c) Divide first_half and second_half with 10. first_half=first_half//10 second_half=second_half//10 (5)if size is even. return rev2*10**mid+rev1 otherwise return (rev2*10**mid+rev1)*10+first_half38

Reverse of no. is 2654_7

Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num1 Input: n (1) Intilize rev1=0, rev2=0 (2) Compute no of digits in given input n and store it in size variable. (3) Divide the number n into two parts i.e first_half and second_half. (4) Loop for i in range(mid): (a) Remainder of first_half by 10 add it to the multiplication of 10 and rev1. rem=first_half%10 rev1=10*rev1+rem (b) Remainder of second_half by 10 add it to the multiplication of 10 and rev2. rem=second_half%10 rev2=10*rev2+rem (c) Divide first_half and second_half with 10. first_half=first_half//10 second_half=second_half//10 (5)if size is even. return rev2*10**mid+rev1 otherwise return (rev2*10**mid+rev1)*10+first_half42Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num1Input: n (1) Intilize rev1=0, rev2=0 (2) Compute no of digits in given input n and store it in size variable. (3) Divide the number n into two parts i.e first_half and second_half. (4) Loop for i in range(mid): (a) Remainder of first_half by 10 add it to the multiplication of 10 and rev1. rem=first_half%10 rev1=10*rev1+rem (b) Remainder of second_half by 10 add it to the multiplication of 10 and rev2. rem=second_half%10 rev2=10*rev2+rem (c) Divide first_half and second_half with 10. first_half=first_half//10 second_half=second_half//10 (5)if size is even. return rev2*10**mid+rev1 otherwise return (rev2*10**mid+rev1)*10+first_half44

Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num6Reverse of no. is 265452 Input: n (1) Intilize rev1=0, rev2=0 (2) Compute no of digits in given input n and store it in size variable. (3) Divide the number n into two parts i.e first_half and second_half. (4) Loop for i in range(mid): (a) Remainder of first_half by 10 add it to the multiplication of 10 and rev1. rem=first_half%10 rev1=10*rev1+rem (b) Remainder of second_half by 10 add it to the multiplication of 10 and rev2. rem=second_half%10 rev2=10*rev2+rem (c) Divide first_half and second_half with 10. first_half=first_half//10 second_half=second_half//10 (5)if size is even. return rev2*10**mid+rev1 otherwise return (rev2*10**mid+rev1)*10+first_half48

Reverse of no. is 2654_2Reverse of no. is 26549 Input: n (1) Intilize rev1=0, rev2=0 (2) Compute no of digits in given input n and store it in size variable. (3) Divide the number n into two parts i.e first_half and second_half. (4) Loop for i in range(mid): (a) Remainder of first_half by 10 add it to the multiplication of 10 and rev1. rem=first_half%10 rev1=10*rev1+rem (b) Remainder of second_half by 10 add it to the multiplication of 10 and rev2. rem=second_half%10 rev2=10*rev2+rem (c) Divide first_half and second_half with 10. first_half=first_half//10 second_half=second_half//10 (5)if size is even. return rev2*10**mid+rev1 otherwise return (rev2*10**mid+rev1)*10+first_half51

Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num6Reverse of no. is 26549 Input: n (1) Intilize rev1=0, rev2=0 (2) Compute no of digits in given input n and store it in size variable. (3) Divide the number n into two parts i.e first_half and second_half. (4) Loop for i in range(mid): (a) Remainder of first_half by 10 add it to the multiplication of 10 and rev1. rem=first_half%10 rev1=10*rev1+rem (b) Remainder of second_half by 10 add it to the multiplication of 10 and rev2. rem=second_half%10 rev2=10*rev2+rem (c) Divide first_half and second_half with 10. first_half=first_half//10 second_half=second_half//10 (5)if size is even. return rev2*10**mid+rev1 otherwise return (rev2*10**mid+rev1)*10+first_half63Input: n (1) Intilize rev1=0, rev2=0 (2) Compute no of digits in given input n and store it in size variable. (3) Divide the number n into two parts i.e first_half and second_half. (4) Loop for i in range(mid): (a) Remainder of first_half by 10 add it to the multiplication of 10 and rev1. rem=first_half%10 rev1=10*rev1+rem (b) Remainder of second_half by 10 add it to the multiplication of 10 and rev2. rem=second_half%10 rev2=10*rev2+rem (c) Divide first_half and second_half with 10. first_half=first_half//10 second_half=second_half//10 (5)if size is even. return rev2*10**mid+rev1 otherwise return (rev2*10**mid+rev1)*10+first_half64Input: n (1) Intilize rev1=0, rev2=0 (2) Compute no of digits in given input n and store it in size variable. (3) Divide the number n into two parts i.e first_half and second_half. (4) Loop for i in range(mid): (a) Remainder of first_half by 10 add it to the multiplication of 10 and rev1. rem=first_half%10 rev1=10*rev1+rem (b) Remainder of second_half by 10 add it to the multiplication of 10 and rev2. rem=second_half%10 rev2=10*rev2+rem (c) Divide first_half and second_half with 10. first_half=first_half//10 second_half=second_half//10 (5)if size is even. return rev2*10**mid+rev1 otherwise return (rev2*10**mid+rev1)*10+first_half65

Reverse of no. is 2654_7

Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num6Input: n (1) Intilize rev1=0, rev2=0 (2) Compute no of digits in given input n and store it in size variable. (3) Divide the number n into two parts i.e first_half and second_half. (4) Loop for i in range(mid): (a) Remainder of first_half by 10 add it to the multiplication of 10 and rev1. rem=first_half%10 rev1=10*rev1+rem (b) Remainder of second_half by 10 add it to the multiplication of 10 and rev2. rem=second_half%10 rev2=10*rev2+rem (c) Divide first_half and second_half with 10. first_half=first_half//10 second_half=second_half//10 (5)if size is even. return rev2*10**mid+rev1 otherwise return (rev2*10**mid+rev1)*10+first_half74

Reverse of no. is 2654_7

Jawa

Input: n (1) Intilize rev1=0, rev2=0 (2) Compute no of digits in given input n and store it in size variable. (3) Divide the number n into two parts i.e first_half and second_half. (4) Loop for i in range(mid): (a) Remainder of first_half by 10 add it to the multiplication of 10 and rev1. rem=first_half%10 rev1=10*rev1+rem (b) Remainder of second_half by 10 add it to the multiplication of 10 and rev2. rem=second_half%10 rev2=10*rev2+rem (c) Divide first_half and second_half with 10. first_half=first_half//10 second_half=second_half//10 (5)if size is even. return rev2*10**mid+rev1 otherwise return (rev2*10**mid+rev1)*10+first_half79 Input: n (1) Intilize rev1=0, rev2=0 (2) Compute no of digits in given input n and store it in size variable. (3) Divide the number n into two parts i.e first_half and second_half. (4) Loop for i in range(mid): (a) Remainder of first_half by 10 add it to the multiplication of 10 and rev1. rem=first_half%10 rev1=10*rev1+rem (b) Remainder of second_half by 10 add it to the multiplication of 10 and rev2. rem=second_half%10 rev2=10*rev2+rem (c) Divide first_half and second_half with 10. first_half=first_half//10 second_half=second_half//10 (5)if size is even. return rev2*10**mid+rev1 otherwise return (rev2*10**mid+rev1)*10+first_half80

#include <stdio.h>8 #include <stdio.h>9

Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num6Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num42 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num05 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num1 Input: n (1) Intilize rev1=0, rev2=0 (2) Compute no of digits in given input n and store it in size variable. (3) Divide the number n into two parts i.e first_half and second_half. (4) Loop for i in range(mid): (a) Remainder of first_half by 10 add it to the multiplication of 10 and rev1. rem=first_half%10 rev1=10*rev1+rem (b) Remainder of second_half by 10 add it to the multiplication of 10 and rev2. rem=second_half%10 rev2=10*rev2+rem (c) Divide first_half and second_half with 10. first_half=first_half//10 second_half=second_half//10 (5)if size is even. return rev2*10**mid+rev1 otherwise return (rev2*10**mid+rev1)*10+first_half29Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num1 Input: n (1) Intilize rev1=0, rev2=0 (2) Compute no of digits in given input n and store it in size variable. (3) Divide the number n into two parts i.e first_half and second_half. (4) Loop for i in range(mid): (a) Remainder of first_half by 10 add it to the multiplication of 10 and rev1. rem=first_half%10 rev1=10*rev1+rem (b) Remainder of second_half by 10 add it to the multiplication of 10 and rev2. rem=second_half%10 rev2=10*rev2+rem (c) Divide first_half and second_half with 10. first_half=first_half//10 second_half=second_half//10 (5)if size is even. return rev2*10**mid+rev1 otherwise return (rev2*10**mid+rev1)*10+first_half31

Reverse of no. is 26542Reverse of no. is 26549 6543211Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num1Original no: 1234 Reversed no: 43210065432151Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num16

Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num6Input: n (1) Intilize rev1=0, rev2=0 (2) Compute no of digits in given input n and store it in size variable. (3) Divide the number n into two parts i.e first_half and second_half. (4) Loop for i in range(mid): (a) Remainder of first_half by 10 add it to the multiplication of 10 and rev1. rem=first_half%10 rev1=10*rev1+rem (b) Remainder of second_half by 10 add it to the multiplication of 10 and rev2. rem=second_half%10 rev2=10*rev2+rem (c) Divide first_half and second_half with 10. first_half=first_half//10 second_half=second_half//10 (5)if size is even. return rev2*10**mid+rev1 otherwise return (rev2*10**mid+rev1)*10+first_half40

________3______6Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num42 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num05 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num1 Input: n (1) Intilize rev1=0, rev2=0 (2) Compute no of digits in given input n and store it in size variable. (3) Divide the number n into two parts i.e first_half and second_half. (4) Loop for i in range(mid): (a) Remainder of first_half by 10 add it to the multiplication of 10 and rev1. rem=first_half%10 rev1=10*rev1+rem (b) Remainder of second_half by 10 add it to the multiplication of 10 and rev2. rem=second_half%10 rev2=10*rev2+rem (c) Divide first_half and second_half with 10. first_half=first_half//10 second_half=second_half//10 (5)if size is even. return rev2*10**mid+rev1 otherwise return (rev2*10**mid+rev1)*10+first_half42Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num1 Input: n (1) Intilize rev1=0, rev2=0 (2) Compute no of digits in given input n and store it in size variable. (3) Divide the number n into two parts i.e first_half and second_half. (4) Loop for i in range(mid): (a) Remainder of first_half by 10 add it to the multiplication of 10 and rev1. rem=first_half%10 rev1=10*rev1+rem (b) Remainder of second_half by 10 add it to the multiplication of 10 and rev2. rem=second_half%10 rev2=10*rev2+rem (c) Divide first_half and second_half with 10. first_half=first_half//10 second_half=second_half//10 (5)if size is even. return rev2*10**mid+rev1 otherwise return (rev2*10**mid+rev1)*10+first_half31

Reverse of no. is 26542Reverse of no. is 265452 Original no: 1234 Reversed no: 432118Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num24 Original no: 1234 Reversed no: 432120

Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num22Reverse of no. is 26549 Input: n (1) Intilize rev1=0, rev2=0 (2) Compute no of digits in given input n and store it in size variable. (3) Divide the number n into two parts i.e first_half and second_half. (4) Loop for i in range(mid): (a) Remainder of first_half by 10 add it to the multiplication of 10 and rev1. rem=first_half%10 rev1=10*rev1+rem (b) Remainder of second_half by 10 add it to the multiplication of 10 and rev2. rem=second_half%10 rev2=10*rev2+rem (c) Divide first_half and second_half with 10. first_half=first_half//10 second_half=second_half//10 (5)if size is even. return rev2*10**mid+rev1 otherwise return (rev2*10**mid+rev1)*10+first_half51

Reverse of no. is 26542Reverse of no. is 26547

Reverse of no. is 26542Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num1 Original no: 1234 Reversed no: 432128Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num24Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num16

Reverse of no. is 26542Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num1 Original no: 1234 Reversed no: 432133Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num24Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num16

Reverse of no. is 26542Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num1 Input: n (1) Intilize rev1=0, rev2=0 (2) Compute no of digits in given input n and store it in size variable. (3) Divide the number n into two parts i.e first_half and second_half. (4) Loop for i in range(mid): (a) Remainder of first_half by 10 add it to the multiplication of 10 and rev1. rem=first_half%10 rev1=10*rev1+rem (b) Remainder of second_half by 10 add it to the multiplication of 10 and rev2. rem=second_half%10 rev2=10*rev2+rem (c) Divide first_half and second_half with 10. first_half=first_half//10 second_half=second_half//10 (5)if size is even. return rev2*10**mid+rev1 otherwise return (rev2*10**mid+rev1)*10+first_half60

Reverse of no. is 26542Reverse of no. is 26549 Original no: 1234 Reversed no: 432141Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num1Original no: 1234 Reversed no: 432143Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num24Original no: 1234 Reversed no: 432145

Reverse of no. is 26542Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num1 Original no: 1234 Reversed no: 432159Original no: 1234 Reversed no: 432160Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num16

Reverse of no. is 26542Original no: 1234 Reversed no: 432163

Reverse of no. is 2654_7

Original no: 1234 Reversed no: 4321_67

Python3

Original no: 1234 Reversed no: 4321_68

Reverse of no. is 12345615 Original no: 1234 Reversed no: 432170

Original no: 1234 Reversed no: 432171Original no: 1234 Reversed no: 432172

Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num6Reverse of no. is 26549 Original no: 1234 Reversed no: 4321756543211Original no: 1234 Reversed no: 432177Original no: 1234 Reversed no: 432178

Original no: 1234 Reversed no: 4321_79

Reverse of no. is 12345615 Original no: 1234 Reversed no: 432181

________3______6Reverse of no. is 265452 Original no: 1234 Reversed no: 432184Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num77 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num24Reverse of no. is 265458Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num64Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num64 Original no: 1234 Reversed no: 432190

Reverse of no. is 26542Reverse of no. is 26549 Original no: 1234 Reversed no: 432193

________3______6Original no: 1234 Reversed no: 432195Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num64 Original no: 1234 Reversed no: 432193________3______77 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num24

________3______6Reverse of no. is 26549 Original no: 1234 Reversed no: 432195Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num83________163______646543211Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num24Original no: 1234 Reversed no: 432145Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num85 #include <stdio.h>20

Reverse of no. is 12345615 #include <stdio.h>22

#include <stdio.h>_30

Reverse of no. is 2654_52 #include <stdio.h>32Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num64Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num64 #include <stdio.h>35#include <stdio.h>36

C#

Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num6Input: n (1) Intilize rev1=0, rev2=0 (2) Compute no of digits in given input n and store it in size variable. (3) Divide the number n into two parts i.e first_half and second_half. (4) Loop for i in range(mid): (a) Remainder of first_half by 10 add it to the multiplication of 10 and rev1. rem=first_half%10 rev1=10*rev1+rem (b) Remainder of second_half by 10 add it to the multiplication of 10 and rev2. rem=second_half%10 rev2=10*rev2+rem (c) Divide first_half and second_half with 10. first_half=first_half//10 second_half=second_half//10 (5)if size is even. return rev2*10**mid+rev1 otherwise return (rev2*10**mid+rev1)*10+first_half40

Reverse of no. is 2654_2Reverse of no. is 265452 #include <stdio.h>77

Reverse of no. is 2654_2Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num1 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num010

Reverse of no. is 2654_7

Javascript

Reverse of no. is 2654_02

Reverse of no. is 2654_85

Reverse of no. is 2654_06

Reverse of no. is 2654_55 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num021

Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num6Input: n (1) Intilize rev1=0, rev2=0 (2) Compute no of digits in given input n and store it in size variable. (3) Divide the number n into two parts i.e first_half and second_half. (4) Loop for i in range(mid): (a) Remainder of first_half by 10 add it to the multiplication of 10 and rev1. rem=first_half%10 rev1=10*rev1+rem (b) Remainder of second_half by 10 add it to the multiplication of 10 and rev2. rem=second_half%10 rev2=10*rev2+rem (c) Divide first_half and second_half with 10. first_half=first_half//10 second_half=second_half//10 (5)if size is even. return rev2*10**mid+rev1 otherwise return (rev2*10**mid+rev1)*10+first_half34

Reverse of no. is 2654_7

Reverse of no. is 2654_2Reverse of no. is 26549 #include <stdio.h>80

Reverse of no. is 2654_7

Reverse of no. is 2654_34

PHP

Reverse of no. is 2654_52

Reverse of no. is 2654_23

Reverse of no. is 2654_06

Reverse of no. is 2654_55 #include <stdio.h>51Reverse of no. is 265457Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num059

Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num6Input: n (1) Intilize rev1=0, rev2=0 (2) Compute no of digits in given input n and store it in size variable. (3) Divide the number n into two parts i.e first_half and second_half. (4) Loop for i in range(mid): (a) Remainder of first_half by 10 add it to the multiplication of 10 and rev1. rem=first_half%10 rev1=10*rev1+rem (b) Remainder of second_half by 10 add it to the multiplication of 10 and rev2. rem=second_half%10 rev2=10*rev2+rem (c) Divide first_half and second_half with 10. first_half=first_half//10 second_half=second_half//10 (5)if size is even. return rev2*10**mid+rev1 otherwise return (rev2*10**mid+rev1)*10+first_half34

Reverse of no. is 2654_7

Reverse of no. is 2654_55 Input: n (1) Intilize rev1=0, rev2=0 (2) Compute no of digits in given input n and store it in size variable. (3) Divide the number n into two parts i.e first_half and second_half. (4) Loop for i in range(mid): (a) Remainder of first_half by 10 add it to the multiplication of 10 and rev1. rem=first_half%10 rev1=10*rev1+rem (b) Remainder of second_half by 10 add it to the multiplication of 10 and rev2. rem=second_half%10 rev2=10*rev2+rem (c) Divide first_half and second_half with 10. first_half=first_half//10 second_half=second_half//10 (5)if size is even. return rev2*10**mid+rev1 otherwise return (rev2*10**mid+rev1)*10+first_half42Reverse of no. is 265457Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num059

________3______6Reverse of no. is 265452 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num075Reverse of no. is 265457 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num077Reverse of no. is 265457Reverse of no. is 265458

Reverse of no. is 2654_2Reverse of no. is 26549 Reverse of no. is 265457Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num16

________3______6Reverse of no. is 26549 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num085 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num102Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num095Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num104Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num090Reverse of no. is 265499

Reverse of no. is 2654_7

Reverse of no. is 2654_90

Reverse of no. is 2654_57 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num110

Reverse of no. is 2654_93 Input: n (1) Intilize rev1=0, rev2=0 (2) Compute no of digits in given input n and store it in size variable. (3) Divide the number n into two parts i.e first_half and second_half. (4) Loop for i in range(mid): (a) Remainder of first_half by 10 add it to the multiplication of 10 and rev1. rem=first_half%10 rev1=10*rev1+rem (b) Remainder of second_half by 10 add it to the multiplication of 10 and rev2. rem=second_half%10 rev2=10*rev2+rem (c) Divide first_half and second_half with 10. first_half=first_half//10 second_half=second_half//10 (5)if size is even. return rev2*10**mid+rev1 otherwise return (rev2*10**mid+rev1)*10+first_half42Reverse of no. is 265457Reverse of no. is 265499

Reverse of no. is 2654_01

Keluaran654321

Kompleksitas Waktu. O(log10n) di mana n adalah nomor input yang diberikan
Ruang Bantu. O(log10n) untuk ruang tumpukan rekursif

Menggunakan String di java

Kami akan mengonversi angka menjadi string menggunakan StringBuffer setelah ini, kami akan membalikkan string itu menggunakan metode reverse()

kasus sudut

Memasukkan. 32100
Jadi untuk input di atas jika kita mencoba menyelesaikannya dengan membalik string, maka outputnya adalah 00123
Jadi untuk menghadapi situasi ini kita perlu mengubah string menjadi integer sehingga output kita menjadi 123

C

#include <stdio.h>

Reverse of no. is 2654_7

Reverse of no. is 26542Reverse of no. is 265452 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num170Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num171Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num21

Reverse of no. is 26542Reverse of no. is 26547

Reverse of no. is 26542Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num178 Reverse of no. is 265452 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num170Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num181Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num21

Reverse of no. is 26542Reverse of no. is 26547

Reverse of no. is 2654_7

C++

Reverse of no. is 1234561

Reverse of no. is 1234562 Reverse of no. is 1234563 Reverse of no. is 1234564

Reverse of no. is 2654_7

Jawa

#include <stdio.h>_7

Reverse of no. is 26542________3______205

Reverse of no. is 2654_2Reverse of no. is 26549 #include <stdio.h>80

Reverse of no. is 26542Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num1 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num50Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num51Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num16

Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num56Input: n (1) Intilize rev1=0, rev2=0 (2) Compute no of digits in given input n and store it in size variable. (3) Divide the number n into two parts i.e first_half and second_half. (4) Loop for i in range(mid): (a) Remainder of first_half by 10 add it to the multiplication of 10 and rev1. rem=first_half%10 rev1=10*rev1+rem (b) Remainder of second_half by 10 add it to the multiplication of 10 and rev2. rem=second_half%10 rev2=10*rev2+rem (c) Divide first_half and second_half with 10. first_half=first_half//10 second_half=second_half//10 (5)if size is even. return rev2*10**mid+rev1 otherwise return (rev2*10**mid+rev1)*10+first_half18

Reverse of no. is 2654_7

Python3

Reverse of no. is 12345615 Reverse of no. is 12345616

________3______6Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num946543211Original no: 1234 Reversed no: 43216Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num391

C#

Reverse of no. is 2654_2Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num125Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num411

Reverse of no. is 26542Reverse of no. is 26549 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num304 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num347________3______418

Reverse of no. is 26542________3______205

Reverse of no. is 2654_2Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num347 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num433

Reverse of no. is 26542Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num50Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num1Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num443

Reverse of no. is 2654_2Reverse of no. is 26549 #include <stdio.h>80

Reverse of no. is 26542Reverse of no. is 265444Original no: 1234 Reversed no: 43216

Reverse of no. is 2654_46Input: n (1) Intilize rev1=0, rev2=0 (2) Compute no of digits in given input n and store it in size variable. (3) Divide the number n into two parts i.e first_half and second_half. (4) Loop for i in range(mid): (a) Remainder of first_half by 10 add it to the multiplication of 10 and rev1. rem=first_half%10 rev1=10*rev1+rem (b) Remainder of second_half by 10 add it to the multiplication of 10 and rev2. rem=second_half%10 rev2=10*rev2+rem (c) Divide first_half and second_half with 10. first_half=first_half//10 second_half=second_half//10 (5)if size is even. return rev2*10**mid+rev1 otherwise return (rev2*10**mid+rev1)*10+first_half18

Reverse of no. is 2654_7

Javascript

Reverse of no. is 2654_02

Reverse of no. is 26542________3______205

Reverse of no. is 2654_2

Reverse of no. is 2654_2Reverse of no. is 26549 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num498

Reverse of no. is 2654_90

Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num6Input: n (1) Intilize rev1=0, rev2=0 (2) Compute no of digits in given input n and store it in size variable. (3) Divide the number n into two parts i.e first_half and second_half. (4) Loop for i in range(mid): (a) Remainder of first_half by 10 add it to the multiplication of 10 and rev1. rem=first_half%10 rev1=10*rev1+rem (b) Remainder of second_half by 10 add it to the multiplication of 10 and rev2. rem=second_half%10 rev2=10*rev2+rem (c) Divide first_half and second_half with 10. first_half=first_half//10 second_half=second_half//10 (5)if size is even. return rev2*10**mid+rev1 otherwise return (rev2*10**mid+rev1)*10+first_half15Original no: 1234 Reversed no: 43216

Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num56Input: n (1) Intilize rev1=0, rev2=0 (2) Compute no of digits in given input n and store it in size variable. (3) Divide the number n into two parts i.e first_half and second_half. (4) Loop for i in range(mid): (a) Remainder of first_half by 10 add it to the multiplication of 10 and rev1. rem=first_half%10 rev1=10*rev1+rem (b) Remainder of second_half by 10 add it to the multiplication of 10 and rev2. rem=second_half%10 rev2=10*rev2+rem (c) Divide first_half and second_half with 10. first_half=first_half//10 second_half=second_half//10 (5)if size is even. return rev2*10**mid+rev1 otherwise return (rev2*10**mid+rev1)*10+first_half18

Reverse of no. is 2654_34

PHP

Reverse of no. is 2654_52

Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num6Reverse of no. is 26549 Reverse of no. is 265457Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num16

Reverse of no. is 2654_7

Reverse of no. is 2654_93 Original no: 1234 Reversed no: 43216Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num554Reverse of no. is 265457Reverse of no. is 265499

Reverse of no. is 2654_01

KeluaranReverse of no. is 123456_

Kompleksitas Waktu. O(log10n) di mana n adalah nomor masukan
Ruang Bantu. O(1)

Membalikkan digit bilangan bulat dengan penanganan luapan

Perhatikan bahwa program di atas tidak mempertimbangkan angka nol di depan. Misalnya untuk 100 program akan mencetak 1. Jika Anda ingin mencetak 001 lihat komentar ini dari Maheshwar

Menggunakan Mengiris dengan Python

C++

Reverse of no. is 1234561

Reverse of no. is 1234562 Reverse of no. is 1234563 Reverse of no. is 1234564

Reverse of no. is 2654_7

Reverse of no. is 2654_90

Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num6Original no: 1234 Reversed no: 43215Original no: 1234 Reversed no: 43216 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num597

Reverse of no. is 2654_7

Jawa

#include <stdio.h>8 #include <stdio.h>9

Reverse of no. is 26542________3______205

Reverse of no. is 26542Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num623Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num304 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num625

Reverse of no. is 2654_2Reverse of no. is 26549 #include <stdio.h>80

Reverse of no. is 26542Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num1 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num50Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num51Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num16

Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num56Input: n (1) Intilize rev1=0, rev2=0 (2) Compute no of digits in given input n and store it in size variable. (3) Divide the number n into two parts i.e first_half and second_half. (4) Loop for i in range(mid): (a) Remainder of first_half by 10 add it to the multiplication of 10 and rev1. rem=first_half%10 rev1=10*rev1+rem (b) Remainder of second_half by 10 add it to the multiplication of 10 and rev2. rem=second_half%10 rev2=10*rev2+rem (c) Divide first_half and second_half with 10. first_half=first_half//10 second_half=second_half//10 (5)if size is even. return rev2*10**mid+rev1 otherwise return (rev2*10**mid+rev1)*10+first_half18

Reverse of no. is 2654_7

Python3

Reverse of no. is 12345615 Reverse of no. is 12345616

________3______6Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num946543211Original no: 1234 Reversed no: 43216Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num391

C#

#include <stdio.h>8 #include <stdio.h>9

Reverse of no. is 26542________3______205

Reverse of no. is 2654_2Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num125Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num731

Reverse of no. is 26542Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num623Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num304 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num347Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num738

Reverse of no. is 2654_2Reverse of no. is 26549 #include <stdio.h>80

Reverse of no. is 26542Reverse of no. is 265444Original no: 1234 Reversed no: 43216

Reverse of no. is 2654_7

Javascript

Original no: 1234 Reversed no: 432171Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num779Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num485Reverse of no. is 265499

Original no: 1234 Reversed no: 4321_71

Original no: 1234 Reversed no: 432171Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num790Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num485Reverse of no. is 12345634

Original no: 1234 Reversed no: 4321_71Reverse of no. is 26549 #include <stdio.h>80

Reverse of no. is 2654_7

Reverse of no. is 2654_90

Reverse of no. is 2654_04

Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num801Original no: 1234 Reversed no: 43216 Input: n (1) Intilize rev1=0, rev2=0 (2) Compute no of digits in given input n and store it in size variable. (3) Divide the number n into two parts i.e first_half and second_half. (4) Loop for i in range(mid): (a) Remainder of first_half by 10 add it to the multiplication of 10 and rev1. rem=first_half%10 rev1=10*rev1+rem (b) Remainder of second_half by 10 add it to the multiplication of 10 and rev2. rem=second_half%10 rev2=10*rev2+rem (c) Divide first_half and second_half with 10. first_half=first_half//10 second_half=second_half//10 (5)if size is even. return rev2*10**mid+rev1 otherwise return (rev2*10**mid+rev1)*10+first_half18

KeluaranReverse of no. is 2654_

Kompleksitas Waktu. O(n), di mana n adalah nomor masukan
Ruang Bantu. O(1)

Coba ekstensi dari fungsi di atas yang seharusnya juga berfungsi untuk angka titik-mengambang

Memecah dan menaklukkan

Algoritma

Python3

Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num831Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num64 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num63Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num91________3______91Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num24Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num83Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num83Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num824

Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num840Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num64 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num63Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num77 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num24Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num83Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num83Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num824

Reverse of no. is 265452Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num904Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num77 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num829 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num64Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num64 Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num15Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num72

Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num6Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num946543211Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num914Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num915Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num83Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num24Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num83Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num83________3______824Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num85______23

Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num6Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num946543211Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num914Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num929Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num83Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num24Input: num (1) Initialize rev_num = 0 (2) Loop while num > 0 (a) Multiply rev_num by 10 and add remainder of num divide by 10 to rev_num rev_num = rev_num*10 + num%10; (b) Divide num by 10 (3) Return rev_num_______________________________________________________________________________________________________________________________________________________________________________________________

KeluaranOriginal no: 1234 Reversed no: 4321

Kompleksitas Waktu. O(log10(setengah no digit dalam angka)), angka menunjukkan input yang diberikan
Ruang Bantu. O(1)

menggunakan membagi dan menaklukkan

Kami akan membagi angka yang diberikan menjadi dua bagian setelah ini, membalikkan babak pertama dan babak kedua dan menggabungkannya

Bagaimana Anda membalikkan angka dengan Python?

Algoritma .

Bilangan Bulat Masukan. nomor

(1) Inisialisasi variabel revs_number = 0

(2) Ulangi angka sementara > 0

(a) Kalikan revs_number dengan 10 dan tambahkan sisa angka

bagi dengan 10 menjadi revs_number

revs_number = revs_number*10 + angka%10;

(b) Bagi bilangan dengan 10

(3) Kembalikan revs_number

Bagaimana cara membalikkan angka dengan Python tanpa menggunakan fungsi balik?

Untuk membalikkan daftar tanpa menggunakan fungsi built-in reverse(), kita menggunakan Slicing Operator . Operator pengiris adalah metode lain yang digunakan untuk membalikkan elemen data.

Bagaimana Anda membalikkan angka dengan Python tanpa loop?

Logika untuk Bilangan Terbalik dengan Python .

Nomor input dapat dibaca dengan menggunakan metode input() atau raw_input()

Selanjutnya, periksa apakah nilai yang dimasukkan adalah bilangan bulat atau tidak

Sekarang periksa apakah bilangan bulat yang diberikan lebih besar dari 0 atau tidak

Buat variabel yang disebut terbalik dan inisialisasi nilai variabel dengan 0

Bagaimana Anda membuat angka mundur?

Cara membalik angka secara matematis. .

Langkah 1 — Isolasi digit terakhir dalam angka. lastDigit = angka % 10. Operator modulo (%) mengembalikan sisa pembagian. .

Langkah 2 — Tambahkan lastDigit untuk mundur. mundur = (membalikkan * 10) + lastDigit. .

Langkah 3-Hapus digit terakhir dari angka. angka = angka / 10

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