5. Longest Palindromic Substring

题目：

Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.

Example:

Input: "babad"Output: "bab"Note: "aba" is also a valid answer.

Example:

Input: "cbbd"Output: "bb"

思路：

本题一开始假设最大回文子字符串长度为Max=s.size()，然后判断是否符合要求，不符合Max--，继续判断，知道发现符合要求的子字符串就直接返回子字符串，但是TimeLimit,但是思路可以参考，算是一种贪心的算法

代码：

class Solution {public:    string longestPalindrome(string s) {        int max = s.size();        string str = s.substr(0,1),sub_str;        int i = 0;        while(1)        {                        for(int j = i;j<=s.size()-max;j++)            {                sub_str = s.substr(j,max);                                if(judge(sub_str))                {                     return sub_str;                }             }            max--;            i=0;            if(max==1)                break;        }    return str;        }private:    bool judge(string s)    {        string::iterator begin = s.begin();            string::iterator end = s.end();          end--;          for(;begin<end;begin++,end--)          {              if(*begin!=*end)              {                flag = end                return false;             }                         }                    return true;      }};

思路：

第二种思路直接采用官方解答方法：Expand Around Center
代码：

class Solution {public:    string longestPalindrome(string s) {    int start = 0, end = 0;    for (int i = 0; i < s.length(); i++) {        int len1 = expandAroundCenter(s, i, i);        int len2 = expandAroundCenter(s, i, i + 1);        int len = max(len1, len2);        if (len > end - start) {            start = i - (len - 1) / 2;            end = i + len / 2;        }    }    return s.substr(start, end-start+1);            }private:    int expandAroundCenter(string s,int left, int right){    int L = left, R = right;    while (L >= 0 && R < s.length() && s[L] == s[R]) {        L--;        R++;    }    return R-L-1;    }};