leetcode--Palindrome Partitioning

Given a string s, partition s such that every substring of the partition is a palindrome.

Return all possible palindrome partitioning of s.

For example, given s = "aab",
Return

  [    ["aa","b"],    ["a","a","b"]  ]

[java] view plain copy

1. public class Solution {  
2.     public List<List<String>> partition(String s) {  
3.         ArrayList<List<String>> res = new ArrayList<List<String>>();  
4.         if(s.length()==0) return res;  
5.         boolean flag[][] = new boolean[s.length()][s.length()];  
6.         for(int i=0;i<s.length();i++){  
7.             flag[i][i] = true;  
8.         }  
9.         for(int i=s.length()-1;i>=0;i--){  
10.             for(int j=i+1;j<s.length();j++){               
11.                 if(s.charAt(i)==s.charAt(j)){  
12.                     if(j-i==1){                       
13.                         flag[i][j] = true;  
14.                     }else{  
15.                         flag[i][j] = flag[i+1][j-1];  
16.                     }  
17.                 }else{  
18.                     flag[i][j] = false;  
19.                 }  
20.             }  
21.         }         
22.         dfs(res, flag, new ArrayList<String>(), 0, s);  
23.         return res;  
24.     }  
25.       
26.     void dfs(ArrayList<List<String>> res,boolean[][] flag,List<String> item,int start,String s){  
27.         if(start==flag.length){  
28.             res.add(new ArrayList<String>(item));  
29.         }else{  
30.             for(int i=start+1;i<flag.length+1;i++){  
31.                 if(flag[start][i-1]){  
32.                     item.add(s.substring(start, i));  
33.                     dfs(res, flag, item, i, s);  
34.                     item.remove(item.size()-1);  
35.                 }  
36.             }  
37.         }  
38.     }  
39. } 

原文链接http://blog.csdn.net/crazy__chen/article/details/46554255