LeetCode 9. Palindrome Number (Easy)

题目

 
​Given an integer x, return true if x is palindrome integer.
An integer is a palindrome when it reads the same backward as forward.
For example, 121 is a palindrome while 123 is not.

Example 1:

Input: x = 121
Output: true
Explanation: 121 reads as 121 from left to right and from right to left.

Example 2:

Input: x = -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: x = 10
Output: false
Explanation: Reads 01 from right to left. Therefore it is not a palindrome.

Constraints:
-231 <= x <= 231 - 1

Follow up: Could you solve it without converting the integer to a string?
 

思路

解法（Java）：

由于题目建议不要将x转为字符串处理，这里我们使用一个int变量reverse来记录x从末尾向前翻转过来的数，循环终止条件x > reverse存在两种情况，

x = reverse 两者位数相同(原x表示的数的位数为偶数)

x < reverse reverse位数比x多一位(原x表示的数的位数为奇数)

class Solution {
    public boolean isPalindrome(int x) {
        // 负数，不为0但以0结尾的数
        if(x < 0 || (x % 10 == 0 && x != 0)) return false;
        int reverse = 0;
        while(x > reverse){
            reverse = reverse * 10 + x % 10;
            x /= 10;
        }
        // 偶数位 和 奇数位情况
        return x == reverse || x == reverse / 10;
    }
}