9. Palindrome Number

//compare half of the digits in x, so don't need to deal with overflow.
class Solution {
public:
    bool isPalindrome(int x) {
        if(x<0 || (x!=0 && x%10==0)) return false;
        int sum=0;
        while(x>sum)
        {
            sum = sum*10+x%10;
            x = x/10;
        }
        return (x==sum)||(x==sum/10);
        //x==sum for even digits and x==sum/10 for odd digits
    }
};

class Solution {
public:
    bool isPalindrome(int x) {
        if (x < 0) return false;
        int div = 1;
        while (x / div >= 10) div *= 10;
        while (x > 0) {
            int left = x / div;
            int right = x % 10;
            if (left != right) return false;
            x = (x % div) / 10;
            div /= 100;
        }
        return true;
    }
};

• http://www.cnblogs.com/grandyang/p/4125510.html
• https://discuss.leetcode.com/topic/12820/an-easy-c-8-lines-code-only-reversing-till-half-and-then-compare