Palindrome Number —— Leetcode（再做一遍）

最新推荐文章于 2020-02-14 13:15:19 发布

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本文介绍了一种不使用额外空间判断整数是否为回文的方法，避免了整数反转时可能产生的溢出问题，并提供了一个具体的C++实现案例。

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

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.

首先想到的大多数是反转整数或用字符串处理，但注意到“反转的边界问题”和“字符串的extra space问题”，我们考虑用单一的运算，找出最前位和最后位来判断，这里有具体方法： http://articles.leetcode.com/2012/01/palindrome-number.html

下面是我的源码，测试用例卡在“1410110141”上，这是10位整数，已经很接近MAX_INT了，不小心把dev*10溢出造成了错误：

class Solution {
public:
    bool isPalindrome(int x) {
        if(x<0) return false;
        int dev=1;
        while(x/dev >= 10)    // Initial wrong solution: x/(dev*10) != 0; pay attention MAX_INT, 1410110141!!
        {
            dev *= 10;
        }
        
        int l, r;
        while(x != 0)
        {
            l = x/dev;
            r = x%10;
            if(l!=r)    return false;
            x = (x%dev)/10;
            dev /= 100;
        }
        return true;
    }
};