3.5LongestPalindromicSubstring

Notes:
	Given a string S, find the longest palindromic substring in S.
	You may assume that the maximum length of S is 1000, and there exists one unique longest palindromic substring.

以某个元素为中心，分别计算偶数长度的回文最大长度和奇数长度的回文最大长度。时间复杂度O(n^2)，空间O（1）

class Solution {
public:
    string longestPalindrome(string s) {
        const int len = s.size();
        if(len <= 1)return s;
        int start, maxLen = 0;
        for(int i = 1; i < len; i++)
        {
            //寻找以i-1,i为中点偶数长度的回文
            int low = i-1, high = i;
            while(low >= 0 && high < len && s[low] == s[high])
            {
                low--;
                high++;
            }
            if(high - low - 1 > maxLen)
            {
                maxLen = high - low -1;
                start = low + 1;
            }
             
            //寻找以i为中心的奇数长度的回文
            low = i- 1; high = i + 1;
            while(low >= 0 && high < len && s[low] == s[high])
            {
                low--;
                high++;
            }
            if(high - low - 1 > maxLen)
            {
                maxLen = high - low -1;
                start = low + 1;
            }
        }
        return s.substr(start, maxLen);
    }
};