# Problem 7 Reverse Integer

## Language

`JavaScript` `Java`

## Description

Given a 32-bit signed integer, reverse digits of an integer.

## Example

1. 123
2. -123
3. 123

1. 321
2. -321
3. 21

### Explanation

Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−231,  231 − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

## Analysis

The biggest problem is to prevent overflow, so we check the origin `x` and the answer `ans`. If they're bigger than `2^31-1`, print `0` directly.
Besides, we reverse the number `x` by read and traverse as string reversed.

## Solution

`JavaScript`

``````/**
* @param {number} x
* @return {number}
*/
var reverse = function(x) {
if(x === 0) return 0;
Math.abs(x)>(2**31-1)?x=0:x;
let ans='';
let rev = Math.abs(x).toString();
for(var i =0;i<rev.length;i++){
ans += rev[rev.length-i-1]
}
return parseInt(Math.abs(ans)>(2**31-1)?len=0:(x<0?-ans:ans));
};``````

`Java`

``````class Solution {
public int reverse(int x) {
long rev= 0;
while( x != 0){
rev= rev*10 + x % 10;
x= x/10;
if( rev > Integer.MAX_VALUE || rev < Integer.MIN_VALUE)
return 0;
}
return (int) rev;

}
}``````