# LeetCode Reverse Integer

## Description

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

## Constraints

-231 <= x <= 231 - 1

## Solution

class Solution {
public int reverse(int x) {
String num = x + "";
String temp = "";
String plusminus = "";
for (int i = num.length()-1; i >=0; i--) {
if (!isInteger(num.charAt(i) + "")) {
plusminus += num.charAt(i);
}
else {
temp += num.charAt(i);
}
}
if (isInteger(plusminus+temp)) {
}
}

boolean isInteger(String s) {
try {
Integer.parseInt(s);
} catch (NumberFormatException e) {
return false;
}
return true;
}
}

## Feedback

class Solution {
public int reverse(int x) {
int rev = 0;
while (x != 0) {
int pop = x % 10;
x /= 10;
if (rev > Integer.MAX_VALUE/10 || (rev == Integer.MAX_VALUE / 10 && pop > 7)) return 0;
if (rev < Integer.MIN_VALUE/10 || (rev == Integer.MIN_VALUE / 10 && pop < -8)) return 0;
rev = rev * 10 + pop;
}
return rev;
}
}

Source