Longest Palindromic Substring

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.

 

参考：

水中的鱼: [LeetCode] Longest Palindromic Substring 解题报告

 

定义函数
P[i,j] = 字符串区间[i,j]是否为palindrome.

首先找个例子，比如S="abccb",
    S=    a  b  c  c  b
Index = 0  1  2  3  4

P[0,0] =1  //each char is a palindrome
P[0,1] =S[0] == S[1]    , P[1,1] =1 
P[0,2] = S[0] == S[2] && P[1,1], P[1,2] = S[1] == S[2] , P[2,2] = 1
P[0,3] = S[0] == S[3] && P[1,2], P[1,3] = S[1] == S[3] && P[2,2] , P[2,3] =S[2] ==S[3],  P[3,3]=1       
......................
由此就可以推导出规律

P[i,j] = 1  if i ==j
        =  S[i] ==S[j]   if j = i+1
        =  S[i] == S[j] && P[i+1][j-1]  if j>i+1

 

public class Solution {
 
    public String longestPalindrome(String s) {
        int len = s.length();
        boolean[][] p = new boolean[len][len];
        int maxL = 0, start=0, end = 0;
        
        for(int i = 0; i < len; i++){
            for(int j = 0; j<i; j++){
                 // i - j < 2 是两个字符中间只隔一个字符
                if((s.charAt(j) == s.charAt(i)) && (i-j<2 || p[j+1][i-1])){
                     p[j][i] = true;
                }
                if(p[j][i] && maxL<(i-j+1)){
                    maxL = i-j+1;
                 // star 和 end 记录最长字符的index
                    start = j;
                    end = i;
                }
            }
            // 每一个字符本事都是 palindrome
            p[i][i] =true;
        }
         return s.substring(start, end+1);
    }
}    

 注意：s.substring(star, end+1); substring是截到第二个参数的前一个。

　　 只有当P[j][i] == true时 才比较maxL