problem

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

也就是说，给定一个字符串A，和一个字符串数组B，判断A能不能由B中的字符串组成。

solution

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def wordBreak(self, s, wordDict):
        """
        :type s: str
        :type wordDict: List[str]
        :rtype: bool
        """

        if s in wordDict:
            return True

        for i in range(1, len(s)):
            if s[0:i] in wordDict:
                if self.wordBreak(s[i:], wordDict) == True:
                    return True

        return False

第一个原始的解决方案，通过扫描字符串进行递归。

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def wordBreak(self, s, wordDict):
        if not s:
            return True

        dp = [False] * (len(s) + 1)
        dp[0] = True

        for i in range(1, len(s) + 1):
            for j in range(i):
                if dp[j] and s[j:i] in wordDict:
                    dp[i] = True
                    break

        return dp[len(s)]

第一个dp方案，关键在于使用dp数组记录状态，当0-j可以被分割，j-i也可以被分割，那么0-i就可以被分割。

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def wordBreak(self, s, wordDict):
        if not s:
            return True

        dp = [False] * (len(s)+1)
        dp[0] = True

        for i in range(len(s)):
            if not dp[i]:
                continue

            for word in wordDict:
                end = i + len(word)
                if end > len(s):
                    continue

                if dp[end]:
                    continue

                if s[i:end] == word:
                    dp[end] = True

        return dp[len(s)]

一点点优化。

summary

使用dp的关键在于缓存中间的结果，减少时间复杂度。