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())){
res[i+1] = true;
break;
}
str.deleteCharAt(0);
}
}
return res[s.length()];
}
public ArrayList<String> wordBreak(String s, Set<String> dict) {
ArrayList<String> res = new ArrayList<String>();
if(s==null || s.length()==0)
return res;
if(wordBreakcheck(s,dict))
helper(s,dict,0,"",res);
return res;
}
private void helper(String s, Set<String> dict, int start, String item, ArrayList<String> res){
if(start>=s.length()){
res.add(item);
return;
}
StringBuilder str = new StringBuilder();
for(int i=start;i<s.length();i++){
str.append(s.charAt(i));
if(dict.contains(str.toString())){
String newItem = new String();
if(item.length()>0)
newItem = item + " " + str.toString();
else
newItem = str.toString();
helper(s,dict,i+1,newItem,res);
}
}
} Reference: http://blog.csdn.net/linhuanmars/article/details/22452163