7. Reverse Integer

class Solution {
    public int reverse(int x) {
        int sign = x < 0 ? -1 : 1;
        x = Math.abs(x);
        int last = 0;
        while (x > 0) {
            if (Integer.MAX_VALUE / 10 < last || (Integer.MAX_VALUE- x % 10) < last * 10) {
                return 0;
            }
            last = last * 10 + x % 10;
            x = x / 10;
        }
        
        return sign * last;
    }
}

Runtime: 1 ms, faster than 100.00% of Java online submissions for Reverse Integer.
Memory Usage: 36.1 MB, less than 68.85% of Java online submissions for Reverse Integer.

Max Value를 확인 안해서 overflow error이 났었음. 그걸 해결하기 위해 overflow line을 더함.