leetcode:Palindrome Number 回文整数

来源：互联网发布：mac 图片浏览编辑：程序博客网时间：2024/06/05 14:10

Determine whether an integer is a palindrome. Do this without extra space.

click to show spoilers.

Some hints:

Could negative integers be palindromes? (ie, -1)

If you are thinking of converting the integer to string, note the restriction of using extra space.

You could also try reversing an integer. However, if you have solved the problem "Reverse Integer", you know that the reversed integer might overflow. How would you handle such case?

There is a more generic way of solving this problem.

判断一个整数是否是回文数，不能使用额外的空间，也就是说你不能将整数转化成字符串然后再来判断，这里我使用的方法是将整数反转再和原来的整数比较，如果相等的话就是回文整数，另外还需要注意的一点是，负数不属于回文数。

class Solution {public:    bool isPalindrome(int x) {        if(x<0)return false;int reversX = 0;int tempX = x;while (tempX>0){reversX = reversX*10 + tempX%10;tempX = tempX/10;}if(x == reversX)return true;return false;    }};

虽然的我的这种做法AC了，但是实际还有一个问题，就是我数字反转之后万一溢出了呢？

另外一种方法有点类似于判断回文字符串，这种方法不会溢出

class Solution {public:    bool isPalindrome(int x) {        if(x<0)return false;int base = 1;int tempX = x;//note: tempX>=10 not tempX>10while (tempX>=10){base = base *10;tempX = tempX/10;}while (x>0){if (x/base != x%10){return false;}else{x = x%base;x = x/10;base = base/100;}}return true;    }};

0 0