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

Posted 2022-12-10 sloanqin

【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i

		int 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i
			if (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