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leetcode 139. word break

admin 11月 13, 2020 0

139. Word Break

Difficulty:: Medium

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, determine if s can be segmented into a space-separated sequence of one or more dictionary words.

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:

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Input: s = "leetcode", wordDict = ["leet", "code"]
Output: true
Explanation: Return true because "leetcode" can be segmented as "leet code".

Example 2:

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Input: s = "applepenapple", wordDict = ["apple", "pen"]
Output: true
Explanation: Return true because "applepenapple" can be segmented as "apple pen apple".
             Note that you are allowed to reuse a dictionary word.

Example 3:

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

Solution

Language: Java

DFS解法

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class  {
    public boolean wordBreak(String s, List<String> wordDict) {
        return dfsHelper(s, new HashSet(wordDict), 0, new Boolean[s.length()]);
    }

    public boolean dfsHelper(String s, Set<String> wordDict, int start, Boolean[] memo) {
        if (start == s.length()) {
            return true;
        }
        if (memo[start] != null) {
            return memo[start];
        }
        for (int end = start + 1; end <= s.length(); end++) {
            if (wordDict.contains(s.substring(start, end)) && dfsHelper(s, wordDict, end, memo)) {
                return memo[start] = true;
            }
        }
        return memo[start] = false;
    }
}

DP解法

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

55%

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class  {
    public boolean wordBreak(String s, List<String> wordDict) {
        if (s == null || wordDict == null) {
            return false;
        }
        Set<String> wordSet = new HashSet<>(wordDict);
        boolean[] dp = new boolean[s.length() + 1];
        dp[0] = true;
        for (int i = 0; i < s.length(); i++) {
            for (int j = i + 1; j <= s.length(); j++) {
                if (dp[i] && wordDict.contains(s.substring(i, j))) {
                    dp[j] = true;
                }
            }
        }
        return dp[s.length()];
    }
}
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