Java for LeetCode 139 Word Break

Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words.

For example, given
s = "leetcode",
dict = ["leet", "code"].

Return true because "leetcode" can be segmented as "leet code".

解题思路一：

直接暴力枚举会导致TLE，因此，需要记录之前的结果，即可以采用dp的思路，JAVA实现如下：

	static public boolean wordBreak(String s, Set<String> wordDict) {
		boolean[] dp = new boolean[s.length() + 1];
		dp[0] = true;
		for (int i = 1; i < dp.length; i++)
			for (int j = i; j >= 0 && !dp[i]; j--)
				if (wordDict.contains(s.substring(i - j, i)))
					dp[i] = dp[i - j];
		return dp[dp.length - 1];
	}

 解题思路二：

考虑到下题用dp做不出来，暴力枚举肯定TLE，所以可以设置一个unmatch集合来存储s中已经确定无法匹配的子串，从而避免重复检查，JAVA实现如下：

    static public boolean wordBreak(String s, Set<String> dict) {
		return wordBreak(s, dict, new HashSet<String>());
	}

	static public boolean wordBreak(String s, Set<String> dict,
			Set<String> unmatch) {
		for (String prefix : dict) {
			if (s.equals(prefix))
				return true;
			else if (s.startsWith(prefix)) {
				String suffix = s.substring(prefix.length());
				if (!unmatch.contains(suffix)) {
					if (wordBreak(suffix, dict, unmatch))
						return true;
					else
						unmatch.add(suffix);
				}
			}
		}
		return false;
	}