LeetCode(5) Longest Palindromic Substring

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题目为：

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.

使用动态规划，可以将时间复杂度降低到0(n^2)。

动态规划方程式为：

设dp[i][j]=true，表示从i到j的子串是回文子串。

则dp[i][j] = dp[i+1][j-1] if s[i]=s[j]

dp[i][j] = false if s[i]!=s[j]

代码如下：

public class Solution {public String longestPalindrome(String s) {int maxlength = 0;int head =0;if(s==null){return null;}if(s.length()==0){return null;}int n = s.length();boolean [][]dp = new boolean [n+1][n+1];for(int i=0;i<n;i++){for(int j=0;j<n;j++){dp[i][j]=false;}dp[i][i]=true;}maxlength = 1;head = 0;for(int i=0;i<n-1;i++){if(s.charAt(i)==s.charAt(i+1)){dp[i][i+1] = true;if(maxlength<2&&dp[i][i+1]){maxlength = 2;head = i;}}}for(int length = 3; length <= n; length ++){for(int i=0;i+length-1<n;i++){if(s.charAt(i)==s.charAt(i+length-1)){dp[i][i+length-1] = dp[i+1][i+length-2];if(length>maxlength&&dp[i][i+length-1]){maxlength = length;head = i;}}}}return s.substring(head,head+maxlength);    }}

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