[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;}