# Reverse Integer LeetCode Solution

Here, We see Reverse Integer LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

# List of all LeetCode Solution

## Problem Statement

Given a signed 32-bit integer `x`, return `x` with its digits reversed. If reversing `x` causes the value to go outside the signed 32-bit integer range `[-231, 231 - 1]`, then return `0`.

Assume the environment does not allow you to store 64-bit integers (signed or unsigned).

Example 1:
Input: x = 123
Output: 321

Example 2:
Input: x = -123
Output: -321

Example 3:
Input: x = 120
Output: 21

## Reverse Integer LeetCode SolutionC++

``````class Solution {
public:
int reverse(int x) {
long r=0;
while(x){
r=r*10+x%10;
x=x/10;
}
if(r>INT_MAX || r<INT_MIN) return 0;
return int(r);
}
};```Code language: PHP (php)```

## Reverse Integer LeetCode SolutionJava

``````class Solution {
public int reverse(int x) {
long finalNum = 0;
while(x!=0){
int lastDig = x%10;
finalNum += lastDig;
finalNum = finalNum*10;
x= x/10;
}
finalNum = finalNum/10;
if(finalNum > Integer.MAX_VALUE || finalNum<Integer.MIN_VALUE){
return 0;
}
if(x<0){
return (int)(-1*finalNum);
}
return (int)finalNum;
}
}```Code language: PHP (php)```

## Reverse Integer SolutionJavaScript

``````var reverse = function(x) {
const absReversed = Math.abs(x).toString().split('').reverse().join('');
if (absReversed > 2**31) return 0;
return absReversed * Math.sign(x);
};```Code language: JavaScript (javascript)```

## Reverse Integer SolutionPython

``````class Solution(object):
def reverse(self, x):
reverse = 0
sign = -1 if x < 0 else 1
x = abs(x)
while x:
digit = x % 10
reverse = reverse * 10 + digit
x /= 10
result = sign * reverse
if result > 2 ** 31 - 1 or result < -(2 ** 31):
return 0
return result```Code language: HTML, XML (xml)```
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