Leet Code 5 Longest Palindromic Substring

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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.

【算法思路】

1. 最简单的办法就是以i为中心进行扩展回文，分奇数和偶数两种情况。O（n^2）

public class Solution {public String longestPalindrome(String s) {int length = s.length();if(length <= 1)return s;String longest = s.substring(0,1);for(int i = 0; i < length-1; i ++){String p1 = expandAroundCenter(s, i, i);if (p1.length() > longest.length())longest = p1;String p2 = expandAroundCenter(s, i, i+1);if (p2.length() > longest.length())longest = p2;}return longest;}String expandAroundCenter(String s, int c1, int c2) {int left = c1, right = c2;int n = s.length();while (left >= 0 && right <= n-1 && s.charAt(left) == s.charAt(right)) {left--;right++;}return s.substring(left+1, right);}
}

2. O(n)解法。相当巧妙，http://www.leetcode.com/2011/11/longest-palindromic-substring-part-ii.html


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

【CODE】

1:       string longestPalindrome(string s) {  
2:            int len = s.size();  
3:            int P[len][len];  
4:            memset(P, 0, len*len*sizeof(int));  
5:            int maxL=0, start=0, end=0;  
6:            for(int i =0; i< s.size(); i++)  
7:            {  
8:                 for(int j =0; j<i; j++)  
9:                 {  
10:                      P[j][i] = (s[j] == s[i] && (i-j<2 || P[j+1][i-1]));  
11:                      if(P[j][i] && maxL < (i-j+1))  
12:                      {  
13:                           maxL = i-j+1;  
14:                           start = j;  
15:                           end = i;  
16:                      }  
17:                 }  
18:                 P[i][i] =1;  
19:            }  
20:            return s.substr(start, end-start +1);  
21:       }  

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