5. Longest Palindromic Substring

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

==

class Solution {
public:
    string longestPalindrome(string s) {
        /**bool dp[i][j]代表s[i-j]是否回文
        dp[i][j] = true         ;i==j
                 = s[i]==s[j]   ;j=i+1
                 = (s[i]==s[j])&& dp[i+1][j-1]; j>i+1
        要记住,dp[i][j]表示的i-j之间的字符串是不是回文,
        当i==j时,dp[i][i]当然是回文
        当j=i+1时,dp[i][j],要看字符串s[i]和s[j]之间是不是相同的?
        当j>i+1时,dp[i][j]的判定情况,s[i]s[j]和dp[i+1][j-1]共同决定的;

        怎么开始的?
        外层循环j控制,
        内存循环i判断当前能到达的位置是否是回文;
        */
        int n = s.size();
        bool dp[n][n];
        fill_n(&dp[][],n*n,false);///初始化
        int max_len = ;
        int start = ;

        for(int j = ;j<s.size();j++){
            for(int i = ;i<=j;i++){
                if(i==j) dp[i][j] = true;
                else if(j==(i+)) dp[i][j] = s[i]==s[j]?true:false;
                else if(j>(i+)) dp[i][j] = s[i]==s[j]&&dp[i+][j-];
                if(dp[i][j]&&max_len < (j-i+)){
                    max_len = j-i+;
                    start = i;
                }
            }///for
        }
        return s.substr(start,max_len);
    }
};