9. Palindrome Number 验证回文数

题目链接
tag:

• easy

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

Example 1:

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?

思路：
  这道验证回文数字的题如果将数字转为字符串，就变成了验证回文字符串的题，没啥难度了，我们就直接来做follow up吧，不能转为字符串，而是直接对整数进行操作，我们可以利用取整和取余来获得我们想要的数字，比如 1221 这个数字，如果 计算 1221 / 1000， 则可得首位1， 如果 1221 % 10， 则可得到末尾1，进行比较，然后把中间的22取出继续比较。代码如下：

C++ 解法：

class Solution {
public:
    bool isPalindrome(int x) {
        /*
        string s = to_string(x);
        int left=0, right=s.size()-1;
        while(left <= right) {
            if(s[left] != s[right])
                return false;
            ++left;
            --right;
        }
        return true;
        */
        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;
    }
};

Python 解法：

class Solution:
    def isPalindrome(self, x: int) -> bool:
        if x < 0:
            return False
        div = 1
        while x / div >= 10:
            div *= 10
        while x > 0:
            left = x // div
            right = x % 10
            if left != right:
                return False
            x = (x % div) // 10
            div /= 100
        return True