7. Reverse Integer

Remembering the formula:

res = res*10 + x%10 (remainder)
x /=10

notice the boundary

time: O(n)

space: O(1)

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