140. Word Break II

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, add spaces in s to construct a sentence where each word is a valid dictionary word. You may assume the dictionary does not contain duplicate words.

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"].

思路：太难了。Using DFS directly will lead to TLE, so I just used HashMap to save the previous results to prune duplicated branches, as the following:

public List<String> wordBreak(String s, Set<String> wordDict) {
    return DFS(s, wordDict, new HashMap<String, LinkedList<String>>());
}       

// DFS function returns an array including all substrings derived from s.
List<String> DFS(String s, Set<String> wordDict, HashMap<String, LinkedList<String>>map) {
    if (map.containsKey(s)) 
        return map.get(s);

    LinkedList<String>res = new LinkedList<String>();     
    if (s.length() == 0) {
        res.add("");
        return res;
    }               
    for (String word : wordDict) {
        if (s.startsWith(word)) {
            List<String>sublist = DFS(s.substring(word.length()), wordDict, map);
            for (String sub : sublist) 
                res.add(word + (sub.isEmpty() ? "" : " ") + sub);               
        }
    }       
    map.put(s, res);
    return res;
}