Java for LeetCode 139 Word Break

標籤：

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".

解題思路一：

直接暴力枚舉會導致TLE，因此，需要記錄之前的結果，即可以採用dp的思路，JAVA實現如下：

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];}

 解題思路二：

考慮到下題用dp做不出來，暴力枚舉肯定TLE，所以可以設定一個unmatch集合來儲存s中已經確定無法匹配的子串，從而避免重複檢查，JAVA實現如下：

    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;elseunmatch.add(suffix);}}}return false;}

 

Java for LeetCode 139 Word Break