LeetCode Word Break II

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”].
思路分析：这题可以用DFS 搜索递归做，基本是brute force的解法，dfs函数要维护的量包括startIndex，preWords和res，startIndex表示当前进行word break的起点，也就是说前面如果已经被break了，应该排除在外。preWords主要维护目前已经breaked的string前面的部分，后面从startIndex一旦发现新的word可以添加到preWords之后，dfs返回的条件是当startIndex已经越界，也就是>=s.length()了，就把当前得到的word break方案加入res。从题目的实例我们也可以看出，同一个字符串可以有多个word break方案，因此我们从前向后scan 字符串发现第一个可以break的词的时候需要更新preWords进行dfs递归调用。这题是NP问题，时间复杂度为O（2^n）也就是指数级别。这类DFS递归搜索的题目有很多，除了word break，还有八皇后，Sudoku Solver等等，都是这类题目，在面试中很常见，要多加练习。
AC Code：以下是brute force DFS搜索AC的从code。由于LeetCode有一个很长的不能break的测试用例，因此把word break I 判断能否break的函数作为sub routine，先判断一下能否break，如果不能直接返回空容器。

public class Solution {public List<String> wordBreak(String s, Set<String> dict) {ArrayList<String> res = new ArrayList<String>();if(s == null || s.isEmpty() || !wordBreakCanDo(s, dict)){return res;}dfs(s, dict, 0, "", res);return res;}//10:08public void dfs(String s, Set<String> dict, int startIndex, String preWords, ArrayList<String> res){if(startIndex >= s.length()){//return contition is that the startIndex has been out of boundres.add(preWords);return;}for(int i = startIndex; i < s.length(); i++){String curStr = s.substring(startIndex, i+1);if(dict.contains(curStr)){String newSol;if(preWords.length() > 0){newSol = preWords + " " + curStr;} else {newSol = curStr;}dfs(s, dict, i + 1, newSol, res);}}}//1021public boolean wordBreakCanDo(String s, Set<String> dict) {s = "#" + s;boolean[] canSegmented = new boolean[s.length()];canSegmented[0] = true;for(int i = 1; i < s.length(); i++){for(int k = 0; k < i; k++){canSegmented[i] = canSegmented[k] && dict.contains(s.substring(k + 1, i + 1));if(canSegmented[i]) break;}}return canSegmented[s.length() - 1];}
}

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