Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward.

Example 1:

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Input: 121
Output: true
Example 2:

Input: -121
Output: false
Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.
Example 3:

Input: 10
Output: false
Explanation: Reads 01 from right to left. Therefore it is not a palindrome.
Follow up:

Coud you solve it without converting the integer to a string?

class Solution {
    public boolean isPalindrome(int x) {
        if(x<0) return false;
        
        int left; //the first bit of the inteter
        int right; //the last bit of the integer
        int divisor = 1; //divisor of each iteration
        int tmp = x;
        
        //determine the initial value of divisor
        while(x/divisor >= 10){
            divisor *= 10;
        }
        
        //check palindrome
        while(x>0){
            left = x/divisor; //divided by divisor to get the first bit
            right = x%10; //mod 10 to get the last bit
            if(left != right) return false;
            
            x = (x%divisor)/10; //mode divisor to remove the first bit, divided by 10 to remove the last bit
            divisor /= 100;
        }
        return true;
    }
}

数字问题注意：

1. 负数

2. 变换后 以0开始 （本题中，如10101）这种情况，除以divisor = 0，mod 10 = 最后那位。

    有多少个0，就得经过多少次循环，才能使得divisor的长度和被除数相等。在长度不等的时候，没次都循环末尾需为0，才能符合Palindrome的判断。所以该算法仍然可以验证有0开头的情况。

 

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