9. Palindrome Number

• 全部反向可能会overflow
• edge cases: negative number; 100,200...
• 将低位反向构成数字，和高位相比

public boolean isPalindrome(int x) {
    if (x < 0 || (x != 0 && x % 10 == 0)) return false;
    int revLow = 0;
    while (x > revLow) {
        revLow = revLow * 10 + x % 10;
        x = x / 10;
    }
    return x == revLow || x == revLow / 10;
}