palindromic java_LeetCode(java)5. Longest Palindromic Substring

Given a stringS, find the longest palindromic substring inS. You may assume that the maximum length ofSis 1000, and there exists one unique longest palindromic substring.

java

题目描述：找出一个字符串中的最长回文子串。算法

本人思路：ide

相似于网上所写的中心扩展法。从字符串中的每一个位置开始向两边扩展，找到最长的回文。spa

时间复杂度：O(n²)，leetcode经过时间：15ms。code

public class Solution {

public static String longestPalindrome(String s) {

char[] cs = s.toCharArray();

int begin=0,end=0;

int size = s.length();

int out = 0;//中心位置

for(int befor=out,after=out;out

{

//判断是否abcba类回文

for(;befor>0&&after

{

if(cs[befor-1] != cs[after+1])

break;

}

if(befor != after && after-befor>end-begin)

{

end=after;

begin=befor;

}

//判断是否abccba类回文

for(befor = out,after = out;befor>=0&&after

{

if(cs[befor] != cs[after+1])

break;

}

if(befor != after && after-(++befor)>end-begin)

{

end=after;

begin=befor;

}

//判断是否aaaaaa类回文

for(befor=out,after=out;after

{

if(cs[befor] != cs[after+1])

break;

}

if(after!=befor && after-befor>end-begin)

{

end=after;

begin=befor;

out = after;//跳太重复的序列

}

return s.substring(begin, end+1);

}

除了这种解法还有如下几种解法：leetcode

①暴力解决法：遍历字符串中的每一个子串，判断是否回文串，并更新最长回文串。字符串

②动态规划：思想为：若 [ i , j ] 为回文，则 [ i+1, j-1 ]也为回文。string

③Manacher's 算法。it

附上大神制做的PPT(很详细)：http://faculty.utpa.edu/liuy2/algorithmSlides/Longest-Palindromic-Substring.pptxio

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