Palindrome Number Solutions in C++

// Source : https://oj.leetcode.com/problems/palindrome-number/
// Author : Hao Chen
// Date   : 2014-06-18



#include <stdio.h>

class Solution {
public:

    bool isPalindrome(int x) {    
        return isPalindrome1(x);
        return isPalindrome2(x);
    }

    bool isPalindrome1(int x) {
        if (x<0) {
            return false;
        }
        
        int len=1;
        for (len=1; (x/len) >= 10; len*=10 );
        
        while (x != 0 ) {
            int left = x / len;
            int right = x % 10;
            
            if(left!=right){
                return false;
            }
            
            x = (x%len) / 10;
            len /= 100;
        }
        return true;
    }
    
    bool isPalindrome2(int x) {
        return (x>=0 && x == reverse(x));
    }

private:    
    int reverse(int x) {
        int y=0;

        int n;
        while( x!=0 ){
            n = x%10;
            //check if y is overflowed or not. 
            //if it is, then it's not palindrome
            if ( y > INT_MAX/10 - n) return x-1;
            y = y*10 + n;
            x /= 10;
        }
        return y;
    }
};



int main()
{
    Solution s;
    printf("%d is %d\n", 0, s.isPalindrome(0) );
    printf("%d is %d\n", -101, s.isPalindrome(-101) );
    printf("%d is %d\n", 1001, s.isPalindrome(1001) );
    printf("%d is %d\n", 1234321, s.isPalindrome(1234321) );
    printf("%d is %d\n", 2147447412, s.isPalindrome(2147447412) );
    printf("%d is %d\n", 2142, s.isPalindrome(2142) );
}

class Solution {
public:
    bool isPalindrome(int x) {
        if (x < 0) return false;
        long rev{0}, origin{x};
        do {
            rev = rev * 10 + x % 10;
        } while (x /= 10);
        return rev == origin;
    }
};