Word Break II leetcode java

题目：

Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each word is a valid dictionary word.

Return all such possible sentences.

For example, given
s = "catsanddog",
dict = ["cat", "cats", "and", "sand", "dog"].

A solution is ["cats and dog", "cat sand dog"].

题解：

这道题不仅仅是看是不是wordbreak，还需要在此基础上把所有word break的结果保存。

为了把所有可能性都保存，那么就使用DFS方法来解决。DFS主要就是跳的层次不容易看出，我下面就以字符串leetcode字典le l et eet code作为例子画了一张图，大概讲解了如何递回和返回，这样更加有助于理解。

 

代码如下：

 1     public boolean wordBreakcheck(String s, Set<String> dict) {
 2         if(s==null || s.length()==0)
 3             return true;
 4         boolean[] res = new boolean[s.length()+1];
 5         res[0] = true;
 6         for(int i=0;i<s.length();i++){
 7             StringBuilder str = new StringBuilder(s.substring(0,i+1));
 8             for(int j=0;j<=i;j++){
 9                 if(res[j] && dict.contains(str.toString())){
10                     res[i+1] = true;
11                     break;
12                 }
13                 str.deleteCharAt(0);
14             }
15         }
16         return res[s.length()];
17     }
18     
19     public ArrayList<String> wordBreak(String s, Set<String> dict) {  
20         ArrayList<String> res = new ArrayList<String>();  
21         if(s==null || s.length()==0)  
22             return res;
23         if(wordBreakcheck(s,dict))
24             helper(s,dict,0,"",res);  
25         return res;  
26     }  
27     private void helper(String s, Set<String> dict, int start, String item, ArrayList<String> res){  
28         if(start>=s.length()){  
29             res.add(item);  
30             return;  
31         }
32         
33         StringBuilder str = new StringBuilder();  
34         for(int i=start;i<s.length();i++){  
35             str.append(s.charAt(i));  
36             if(dict.contains(str.toString())){  
37                 String newItem = new String();  
38                 if(item.length()>0)
39                     newItem = item + " " + str.toString();
40                 else
41                     newItem = str.toString();
42                 helper(s,dict,i+1,newItem,res);  
43             }  
44         }  
45     }  

 Reference: http://blog.csdn.net/linhuanmars/article/details/22452163