如何反向一个整数的位数（（包括溢出处理）） _数据结构

本文概述

C ++
Java
python
C#
的PHP
C ++
C
Java
Python3
C#

假设输入是32位整数, 则编写程序以反转整数。如果反向整数溢出, 则输出-1作为输出。
让我们看看整数位数反转的简单方法.
C ++

// A simple C program to reverse digits of // an integer. #include < stdio.h> int reversDigits( int num) { int rev_num = 0; while (num > 0) { rev_num = rev_num*10 + num%10; num = num/10; } return rev_num; }/* Driver program to test reversDigits */ int main() { int num = 5896; printf ( "Reverse of no. is %d" , reversDigits(num)); return 0; }

Java

// Java program to reverse a number class GFG { /* Iterative function to reverse digits of num*/ static int reversDigits( int num) { int rev_num = 0 ; while (num > 0 ) { rev_num = rev_num * 10 + num % 10 ; num = num / 10 ; } return rev_num; }// Driver code public static void main (String[] args) { int num = 4562 ; System.out.println( "Reverse of no. is " + reversDigits(num)); } }// This code is contributed by Anant Agarwal.

python

# Python program to reverse a number n = 4562 ; rev = 0while (n > 0 ): a = n % 10 rev = rev * 10 + a n = n / 10print (rev)# This code is contributed by Shariq Raza

// C# program to reverse a number using System; class GFG { // Iterative function to // reverse digits of num static int reversDigits( int num) { int rev_num = 0; while (num > 0) { rev_num = rev_num * 10 + num % 10; num = num / 10; } return rev_num; }// Driver code public static void Main() { int num = 4562; Console.Write( "Reverse of no. is " + reversDigits(num)); } }// This code is contributed by Sam007

的PHP

< ?php // Iterative function to // reverse digits of num function reversDigits( $num ) { $rev_num = 0; while ( $num > 1) { $rev_num = $rev_num * 10 + $num % 10; $num = (int) $num / 10; } return $rev_num ; }// Driver Code $num = 4562; echo "Reverse of no. is " , reversDigits( $num ); // This code is contributed by aj_36 ?>

输出如下：

6985

【如何反向一个整数的位数（（包括溢出处理））】但是, 如果数量很大, 以至于反向溢出, 则输出是一些垃圾值。如果我们使用输入为任意大的数字来运行上面的代码, 请说1000000045, 则输出是一些垃圾值, 例如1105032705或任何其他垃圾值。看到这个用于输出。
如何处理溢出？
想法是存储和的先前值可以存储在变量中, 该变量可以每次检查以查看反向是否溢出。
下面是处理这种情况的实现。
C ++

// C++ program to reverse digits // of a number #include < bits/stdc++.h> using namespace std; /* Iterative function to reverse digits of num*/ int reversDigits( int num) { // Handling negative numbers bool negativeFlag = false ; if (num < 0) { negativeFlag = true ; num = -num ; }int prev_rev_num = 0, rev_num = 0; while (num != 0) { int curr_digit = num % 10; rev_num = (rev_num * 10) + curr_digit; // checking if the reverse overflowed or not. // The values of (rev_num - curr_digit)/10 and // prev_rev_num must be same if there was no // problem. if ((rev_num - curr_digit) / 10 != prev_rev_num) { cout < < "WARNING OVERFLOWED!!!" < < endl; return 0; }prev_rev_num = rev_num; num = num / 10; }return (negativeFlag == true ) ? -rev_num : rev_num; }// Driver Code int main() { int num = 12345; cout < < "Reverse of no. is " < < reversDigits(num) < < endl; num = 1000000045; cout < < "Reverse of no. is " < < reversDigits(num) < < endl; return 0; }// This code is contributed // by Akanksha Rai(Abby_akku)

// C program to reverse digits of a number #include < stdio.h> /* Iterative function to reverse digits of num*/ int reversDigits( int num) { // Handling negative numbers bool negativeFlag = false ; if (num < 0) { negativeFlag = true ; num = -num ; }int prev_rev_num = 0, rev_num = 0; while (num != 0) { int curr_digit = num%10; rev_num = (rev_num*10) + curr_digit; // checking if the reverse overflowed or not. // The values of (rev_num - curr_digit)/10 and // prev_rev_num must be same if there was no // problem. if ((rev_num - curr_digit)/10 != prev_rev_num) { printf ( "WARNING OVERFLOWED!!!\n" ); return 0; }prev_rev_num = rev_num; num = num/10; }return (negativeFlag == true )? -rev_num : rev_num; }/* Driver program to test reverse Digits */ int main() { int num = 12345; printf ( "Reverse of no. is %d\n" , reversDigits(num)); num = 1000000045; printf ( "Reverse of no. is %d\n" , reversDigits(num)); return 0; }

Java

// Java program to reverse digits of a numberclass ReverseDigits { /* Iterative function to reverse digits of num*/ static int reversDigits( int num) { // Handling negative numbers boolean negativeFlag = false ; if (num < 0 ) { negativeFlag = true ; num = -num ; }int prev_rev_num = 0 , rev_num = 0 ; while (num != 0 ) { int curr_digit = num% 10 ; rev_num = (rev_num* 10 ) + curr_digit; // checking if the reverse overflowed or not. // The values of (rev_num - curr_digit)/10 and // prev_rev_num must be same if there was no // problem. if ((rev_num - curr_digit)/ 10 != prev_rev_num) { System.out.println( "WARNING OVERFLOWED!!!" ); return 0 ; }prev_rev_num = rev_num; num = num/ 10 ; }return (negativeFlag == true )? -rev_num : rev_num; }public static void main (String[] args) { int num = 12345 ; System.out.println( "Reverse of no. is " + reversDigits(num)); num = 1000000045 ; System.out.println( "Reverse of no. is " + reversDigits(num)); } }

Python3

# Python program to reverse digits # of a number """ Iterative function to reverse digits of num""" def reversDigits(num): # Handling negative numbers negativeFlag = False if (num < 0 ):negativeFlag = True num = - num prev_rev_num = 0 rev_num = 0 while (num ! = 0 ): curr_digit = num % 10rev_num = (rev_num * 10 ) + curr_digit # checking if the reverse overflowed or not. # The values of (rev_num - curr_digit)/10 and # prev_rev_num must be same if there was no # problem. if (rev_num > = 2147483647 or rev_num < = - 2147483648 ): rev_num = 0 if ((rev_num - curr_digit) / / 10 ! = prev_rev_num):print ( "WARNING OVERFLOWED!!!" ) return 0prev_rev_num = rev_num num = num / / 10return - rev_num if (negativeFlag = = True ) else rev_num # Driver code if __name__ = = "__main__" : num = 12345 print ( "Reverse of no. is " , reversDigits(num)) num = 1000000045 print ( "Reverse of no. is " , reversDigits(num)) # This code is contributed # Shubham Singh(SHUBHAMSINGH10)

// C# program to reverse digits // of a number using System; class GFG {/* Iterative function to reverse digits of num*/ static int reversDigits( int num) { // Handling negative numbers bool negativeFlag = false ; if (num < 0) { negativeFlag = true ; num = -num ; } int prev_rev_num = 0, rev_num = 0; while (num != 0) { int curr_digit = num % 10; rev_num = (rev_num * 10) + curr_digit; // checking if the reverse overflowed // or not. The values of (rev_num - // curr_digit)/10 and prev_rev_num must // be same if there was no problem. if ((rev_num - curr_digit) / 10 != prev_rev_num) { Console.WriteLine( "WARNING OVERFLOWED!!!" ); return 0; } prev_rev_num = rev_num; num = num / 10; } return (negativeFlag == true ) ? -rev_num : rev_num; } // Driver Code static public void Main () { int num = 12345; Console.WriteLine( "Reverse of no. is " + reversDigits(num)); num = 1000000045; Console.WriteLine( "Reverse of no. is " + reversDigits(num)); } }// This code is contributed by ajit

输出如下：