LeetCode(5) Longest Palindromic Substring

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);
}
}