给定一个字符串 s，找到 s 中最长的回文子串。你可以假设 s 的最大长度为 1000。

资质有限，所以我就只好，写了一个非常差的example，而且我这个也不能保证从这个字符串中

返回来的是一个非常的长的回文子串。

class Solution {
public:
    string longestPalindrome(string s) {
        int sizenum = s.size();
        if(sizenum < 2)return NULL;
        //string min_s = NULL;
        int sign_i = 0;
        int sign_j = 0;
        for(int i = 0;i < sizenum;i++){
            for(int j=sizenum - 1;j > i;j--){
                if(s[i] == s[j]){
                    sign_i = i;
                    sign_j = j;
                    while(sign_i != sign_j && 
                    s[sign_i] == s[sign_j] &&
                    sign_i < sign_j){
                        sign_i ++;
                        sign_j --;
                    }
                    if( sign_i == sign_j || sign_i > sign_j){
                    int sign_size = j - i + 1;
                    string min_s = s.substr(i,sign_size);
                        return min_s;
                    }
                }
            }
        }
        return NULL;
    }
};

从leetcode上找了一个非常好的例子

下面贴出来代码：

class Solution {
public:
    string longestPalindrome(string s) {
        int start = 0, end = 0, len = 0, g_start = 0, i = 0, n = s.size();
        while(i<n) {
            start = i, end = i;
            while(i+1<n && s[end] == s[end+1]) end++ ;
            i = end + 1;
            while(start-1>=0 && end+1<n && s[end+1] == s[start-1]) {
                start-- ;
                end++ ;
            }
            if(end - start + 1 > len) {
                len = end - start + 1;
                g_start = start;
            }
        }
        cout<<s.substr(g_start, len);
        return s.substr(g_start, len);
    }
};

class Solution {

public:

    string longestPalindrome(string s) {

        int sizenum = s.size();

        if(sizenum < 2)return NULL;

        //string min_s = NULL;

        int sign_i = 0;

        int sign_j = 0;

        for(int i = 0;i < sizenum;i++){

            for(int j=sizenum - 1;j > i;j--){

                if(s[i] == s[j]){

                    sign_i = i;

                    sign_j = j;

                    while(sign_i != sign_j && 

                    s[sign_i] == s[sign_j] &&

                    sign_i < sign_j){

                        sign_i ++;

                        sign_j --;

                    }

                    if( sign_i == sign_j || sign_i > sign_j){

                    int sign_size = j - i + 1;

                    string min_s = s.substr(i,sign_size);

                        return min_s;

                    }

                }

            }

        }

        return NULL;

    }

};

 

 

https://leetcode-cn.com/problems/longest-palindromic-substring/solution/zui-chang-hui-wen-zi-chuan-c-by-gpe3dbjds1/