9. Palindrome Number

https://leetcode.com/problems/palindrome-number/description/
Solution 1. Only compare half digits (fastest). Corner cases are (x!=0 && x%10 == 0).
    bool isPalindrome(int x) {
        if(x<0 || (x!=0 && x%10==0)) return false;
        int y = 0;
        while(x>y) {
            y = y*10 + (x%10);
            x /= 10;
        }
        return x == y || (x == y/10);
    }
Solution 1.1 Reverse whole number
    bool isPalindrome(int x) {
        if(x<0) return false;
        int a = x;
        int y = 0;
        while(a) {
            y = y*10 + (a%10);
            a /= 10;
        }
        return x == y;
    }
Solution 2. Covert to string and compare it to its reverse.
    bool isPalindrome(int x) {
        if(x<0) return false;
        string s = to_string(x);
        string t = s;
        reverse(begin(t),end(t));
        return s == t;
    }
Solution 2.1 Two pointers for string.
    bool isPalindrome(int x) {
        if(x<0) return false;
        string s = to_string(x);
        for(int i=0, j=s.size()-1;i<j; i++,j--){
            if(s[i] != s[j]) return false;
        }
        return true;
    }