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  • 139. Word Break

    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:

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

    Example 2:

    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:

    Input: s = "catsandog", wordDict = ["cats", "dog", "sand", "and", "cat"]
    Output: false
    class Solution(object):
        def wordBreak(self, s, wordDict):
            """
            :type s: str
            :type wordDict: List[str]
            :rtype: bool
            """
            n = len(s)
            dp = [False] * n
            
            for i in xrange(n):
                for w in wordDict:
                    if w == s[i-len(w)+1:i+1] and (dp[i-len(w)] or i-len(w) == -1):
                        dp[i] =True
            
            return dp[-1]
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  • 原文地址:https://www.cnblogs.com/boluo007/p/12506454.html
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