[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"].

这题跟Word Break I一样，简单的使用递归会超时。结合前一题的解法做剪枝，只对能够被拆解的子字符串进行后续DFS，这样可以有效减少DFS的搜索空间。

可以直接照搬前一版本的代码。

    public List<String> wordBreak(String s, Set<String> dict) {
        List<String> ret = new ArrayList<String>(); 

		// An important pruning here, by resuing the version I.
		// Return an empty collection if no solution exists.
		if (!isBreakable(s, dict)) {
		    return ret;
		}
		
		dfs(s, dict, 0, "", ret);
		return ret;
    }
    
    private boolean isBreakable(String s, Set<String> dict) {
		// dp[i][j] represents the substring that starts at i and ends at j,
		// where j is exclusive.
		// dp[*][0] is wasted.
		boolean[][] dp = new boolean[s.length()][s.length() + 1];

		for (int len = 1; len <= s.length(); ++len) {
			for (int start = 0; start <= s.length() - len; ++start) {
				int end = start + len;
				String substr = s.substring(start, end);
				if (len == 1) {
					dp[start][end] = dict.contains(substr);
				} else {
					if (dict.contains(substr)) {
						dp[start][end] = true;
						continue;
					}

					// Try to partition the word.
					for (int leftLen = 1; leftLen < len; ++leftLen) {
						int leftEnd = start + leftLen;
						if (dp[start][leftEnd] && dp[leftEnd][end]) {
							dp[start][end] = true;
							break;
						}
					}
				}
			}
		}

		return dp[0][s.length()];
	}
    
    private void dfs(String s, Set<String> dict, int start, String sentence, List<String> sentences) {
        if (start == s.length()) {
            // Remove the leading space before adding it to the result collection.
            sentences.add(sentence.substring(1));
            return;
        }
        
        for (int len = 1; start + len <= s.length(); ++len) {
            int end = start + len;
            String str = s.substring(start, end);
            if (dict.contains(str)) {
                dfs(s, dict, end, sentence + " " + str, sentences);
            }
        }
     }