[LeetCode] 9. Palindrome Number

思路:
看当前数字的反数字(reverse)是否和原数字相等

bool isPalindrome(int x) {
    if (x < 0) return false;
    if (! x) return true;
    int n1 = 0, n2 = x;
    while (x) {
        n1 = n1 * 10 + x % 10;
        x /= 10;
    }
    return n1 == n2;
}