[leetcode] 139. Word Break @ python

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原题

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

解法

动态规划, 以 s = “leetcode”, wordDict = [“leet”, “code”] 为例, 初始化dp[i]表示截止至index i是否有字符串在字典中, 那么dp[3]为真, dp[7]为真的条件是dp[3]为真且s[4:7]在字典中. 这样我们查找dp[-1]是否为真即可, 表示整个s能完整的分割为字典中的单词.

代码

class Solution:
    def wordBreak(self, s, wordDict):
        """
        :type s: str
        :type wordDict: List[str]
        :rtype: bool
        """
        dp = [False]*len(s)        
        for i in range(len(s)):
            for j in range(i, len(s)):
                if s[i:j+1] in wordDict and (dp[i-1]== True or i-1 == -1):
                    dp[j] = True
        return dp[-1]