LeetCode——最长连续回文串

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

给出一个字符串S，找到一个最长的连续回文串。你可以假设s的最大长度是1000。

Example:

Input: "babad"

Output: "bab"

Note: "aba" is also a valid answer.

Example:

Input: "cbbd"

Output: "bb"

动态规划：

class Solution {
public:
    string longestPalindrome(string s) {
        if (s.empty()) return "";
        if (s.size() == 1) return s;
        
        int min_start = 0;
        int max_len = 1;
        int n = s.size();
        for (int i = 0; i < n;) {
            int j = i, k = i;
            while (k < n-1 && s[k] == s[k+1]) {//k——k+1表示相等的同一个元素
                k++;                        
            }
            i = k + 1;
            while (k < n-1 && s[k+1] == s[j-1] && j > 0) {//j-1——k+1为对称相等的元素
                ++k;
                --j;
            }
            
            int new_len = k - j + 1;//当前回文串大小
            if (new_len > max_len) {
                min_start = j;
                max_len = new_len;
            }
        }
        return s.substr (min_start, max_len);
    }
};

?Manacher’s Algorithm算法

 用 mx 记录之前计算的最长的回文子串长度所能到达的最后边界，用 id 记录其对应的中心，可以利用回文子串中的回文子串信息。

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);
    }
};