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"]
  ]

列出字符串分解的题基本都是DFS，本题是找出s所有合数的分解，故i=start

在permutation那题中，需要列出所有可能的排列，每次递归时是从i=0开始

 1 public class Solution {
 2     public ArrayList<ArrayList<String>> partition(String s) {
 3         // Start typing your Java solution below
 4         // DO NOT write main() function
 5         ArrayList<ArrayList<String>> result = new ArrayList<ArrayList<String>>();
 6         ArrayList<String> output = new ArrayList<String>();
 7         int depth = 0, len = s.length();
 8         
 9         palinPartition(s, 0, len, output, result);
10         return result;
11     }
12     
13     public void palinPartition(String s, int start, int len, ArrayList<String> output,
14                         ArrayList<ArrayList<String>> result){
15         if(start == len){
16             ArrayList<String> tmp = new ArrayList<String>();
17             tmp.addAll(output);
18             result.add(tmp);
19             return;
20         }
21         
22         for(int i = start; i < len; i++){
23             if(isPalindrome(s, start, i)){
24                 output.add(s.substring(start, i + 1));
25                 palinPartition(s, i + 1, len, output, result);
26                 output.remove(output.size() - 1);
27             }
28         }
29         
30     }
31     
32     public boolean isPalindrome(String s, int start, int end){
33         while(start < end){
34             if(s.charAt(start) != s.charAt(end)){
35                 return false;
36             }
37             start ++;
38             end --;
39         }
40         
41         return true;
42     }
43 }