腾讯//最长回文子串

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。

示例 1：

输入: "babad"
输出: "bab"
注意: "aba"也是一个有效答案。

示例 2：

输入: "cbbd"
输出: "bb"

class Solution {
public:
    string longestPalindrome(string s) {
        if(s.size()<2) return s;
        int n = s.size(),maxLen = 0, start = 0;
        for(int i = 0; i < n - 1; i++){
            searchPalindrome(s,i,i,start,maxLen);
            searchPalindrome(s,i,i+1,start,maxLen);
        }
        return s.substr(start, maxLen);
    }
    void searchPalindrome(string s, int left, int right, int &start, int &maxLen){
        while(left>=0&&right<s.size()&&s[left] == s[right]){
            --left;
            ++right;
        }
        if(maxLen < right - left - 1){
            start = left + 1;
            maxLen = right - left - 1;
        }
    }
};

class Solution {
public:
    string longestPalindrome(string s) {
        if(s.size() < 2)
            return s;
        int n = s.size(),maxLen = 0, start = 0;
        for(int i = 0; i < n;){
            if(n - i <= maxLen/2) break;
            int left = i, right = i;
            while(right < n - 1 && s[right+1] == s[right]) ++right;
            i = right + 1;
            while(right < n - 1 && left > 0 && s[right+1] == s[left-1]){
                ++right;
                --left;
            }
            if(maxLen < right - left + 1){
                maxLen = right - left + 1;
                start = left;
            }
        }
        return s.substr(start, maxLen);
    }
};

class Solution {
public:
    string longestPalindrome(string s) {
        if(s.empty()) return "";
        int dp[s.size()][s.size()] = {0}, left = 0, right = 0, len = 0;
        for(int i = 0; i < s.size(); i++){
            for(int j = 0; j < i; j++){
                dp[j][i] = (s[i]==s[j]&&(i-j<2||dp[j+1][i-1]));
                if(dp[j][i]&&len<i-j+1){
                    len = i-j+1;
                    left = j;
                    right = i;
                }
            }
            dp[i][i] = 1;
        }
        return s.substr(left,right-left+1);
    }
};

class Solution {
public:
    string longestPalindrome(string s) {
        string t = "$#";
        for(int i = 0; i < s.size(); i++){
            t += s[i];
            t += '#';
        }
        int p[t.size()] = {0}, id = 0, mx = 0, resId = 0, resMx = 0;
        for(int i = 0; i < t.size(); i++){
            p[i] = mx > i ? min(p[2*id-i], mx-i) : 1;
            while(t[i + p[i]] == t[i - p[i]]) ++p[i];
            if(mx < i+p[i]){
                mx = i + p[i];
                id = i;
            }
            if(resMx < p[i]){
                resMx = p[i];
                resId = i;
            }
        }
        return s.substr((resId - resMx) / 2, resMx-1);
    }
};