Reverse digits of an integer.

Example1: x = 123, return 321

Example2: x = -123, return -321

My favorite solution

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public class Solution {
	public int reverse(int x) {
        long rev = 0; // possible overflow
        while ( x != 0) { // no need to take care of negative value
            rev = rev * 10 + x % 10; // no need for an array to keep track of digits
            x = x / 10;
            if (rev > Integer.MAX_VALUE || rev < Integer.MIN_VALUE) // check for overflow while reversing
                return 0;
        }
        return (int) rev;
    }
}

My solution:

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public class Solution {
    public int reverse(int x) {
        long rev = 0;
        boolean neg = false;
        
        if (x < 0){
            neg = true;
            x = -x;
        }
        
        while (x > 0){
            rev = rev * 10 + x % 10;
            x /= 10;
        }
        
        if (rev > (long) Integer.MAX_VALUE){return 0;}
        if (neg){rev = -rev;}
        return (int) rev;
    }
}

Most voted solution on Leetcode

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public class Solution {
    public int reverse(int x) {
        int rev = 0;
        
        while (x != 0){
            int newR = rev * 10 + x % 10;
            if ((newR - x % 10) / 10 != rev){return 0;}
            rev = newR;
            x /= 10;
        }
        
        return rev;
    }
}