leetcode--Longest Palindromic Substring--最长回文子串--

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

【算法说明】：针对回文子串有一个很简单效果又很好的算法：manacher算法，大家可以亲切的称之为马拉车算法；

关于该算法的原理，可以参考博客：点击打开链接

【自己的代码】：

#include <iostream>#include <string>#include <queue>#include <memory.h>using namespace std;class Solution {public://add #void preTreat(string &strIn, string &strOut){int i = 0;for (auto iter = strIn.begin(); iter != strIn.end(); ++iter){strOut[i] = '#';strOut[i + 1] = *iter;i = i + 2;}strOut[i] = '#';}void manacher(string &strIn, string &strOut){string strChanged;strChanged.resize(strIn.size() * 2 + 1);preTreat(strIn, strChanged);cout << strChanged << endl;int len = strChanged.size();int *rad = (int *)malloc(sizeof(int)*len);memset(rad, 0, sizeof(int)*len);int maxIndex = 0;//记录目前为止，考察过的字符串中最大的索引值int iOfMaxIndex = 0;//maxIndex 对应的iint ans=0;//记录最长回文子串中心位置的索引编号for (int i = 0; i < len; i++){//这个if语句帮助减少了很大的计算量if (maxIndex >= i){rad[i] = min(rad[2 * iOfMaxIndex - i], maxIndex - i);//cout << "after change,rad[i]: " << rad[i] << endl;}//以该点为中心，向两边扩展，考察最大的半径长度while (((i - rad[i] - 1) >= 0) && ((i + rad[i] + 1) <= len) && (strChanged[i + rad[i] + 1] == strChanged[i - (rad[i] + 1)])){rad[i]++;}//保存新的到最大索引的iif (maxIndex < (i + rad[i])){maxIndex = i + rad[i];iOfMaxIndex = i;}if (rad[ans] < rad[i]){ans = i;}}for (int i = (ans - rad[ans]); i <= ans + rad[ans]; ++i){if (strChanged[i] != '#'){strOut.resize(strOut.size() + 1, strChanged[i]);}}cout << strOut << endl;delete rad;}string longestPalindrome(string s) {string strAns;manacher(s, strAns);return strAns;}};

作者：香蕉麦乐迪--sloanqin--覃元元