我之前竟然没有记这道题……奇怪了嘿……

问题描述

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"

问题分析

有成熟的算法来解这个问题：Manacher's Algorithm 马拉车算法
核心思想是，记录当前回文已经到达的最右边的边界，若新位置在边界内，则可以利用回文的对称性，减少计算。

AC代码

class Solution {
public:
    int min(int a, int b){
        if (a < b)
            return a;
        else
            return b;
    }
    string preprocessing(string s){
        int len = s.size();
        string rst = "^#";
        for (auto c: s){
            rst += c;
            rst += '#';
        }
        rst += '$';
        return rst;
    }
    string longestPalindrome(string s) {
        string rst = preprocessing(s);
        vector<int> p(rst.size(), 1);
        int center = 1;
        int brim = 1;
        int max = 1;
        for (int i = 2; i < rst.size(); i++){
            int t = i < brim ? min(p[2 * center - i], brim - i) : 1;
            int j1 = i + t;
            int j2 = i - t;
            while (rst[j1] == rst[j2]){
                t++;
                j1++;
                j2--;
            }
            p[i] = t;
            if (t > p[max]){
                max = i;
            }
            if (i + t > brim){
                center = i;
                brim = i + t;
            }
        }
        string r = rst.substr(2 * max - (max + p[max] - 1), 2 * p[max] - 1);
        for (auto it = r.begin(); it != r.end();){
            if (*it == '#' || *it == '^' || *it == '$')
                it=r.erase(it);
            else
                it++;
        }
        return r;
    }
};

Runtime: 6 ms, which beats 76.09% of cpp submissions.