題目：Single Number

難度：hard

問題描述：

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. Return all such possible sentences.

Note:

• The same word in the dictionary may be reused multiple times in the segmentation.
• You may assume the dictionary does not contain duplicate words.

Example 1:

Input:
s = "catsanddog"
wordDict = ["cat", "cats", "and", "sand", "dog"]
Output:
[
  "cats and dog",
  "cat sand dog"
]

Example 2:

Input:
s = "pineapplepenapple"
wordDict = ["apple", "pen", "applepen", "pine", "pineapple"]
Output:
 
[
  "pine apple pen apple",
  "pineapple pen apple",
  "pine applepen apple"
]
Explanation: Note that you are allowed to reuse a dictionary word.

Example 3:

Input:
s = "catsandog"
wordDict = ["cats", "dog", "sand", "and", "cat"]
Output:
[]

求解思路：

使用DFS，分割成子問題求解。使用map儲存歷史記錄，減少工作量。

程式碼如下：

//找到所有的分割方案，分割使用空格符
	public List<String> wordBreak2(String s, List<String> wordDict) {
		return DFS(s, wordDict, new HashMap<String, LinkedList<String>>());
    }
	List<String> DFS(String s,List<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;
	}