题目

Given a string s and a dictionary of strings wordDict, return true if s can be segmented into a space-separated sequence of one or more dictionary words.

Note that the same word in the dictionary may be reused multiple times in the segmentation.

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

Constraints:

• 1 <= s.length <= 300
• 1 <= wordDict.length <= 1000
• 1 <= wordDict[i].length <= 20
• s and wordDict[i] consist of only lowercase English letters.
• All the strings of wordDict are unique.

思路

动态规划

代码

python版本：

# dfs，超时
class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        words = Counter(wordDict)

        def dfs(now):
            if now == len(s):
                return True
            nexts = []
            for i in range(now+1, len(s)+1):
                if s[now:i] in words:
                    nexts.append(i)
            return any([dfs(i) for i in nexts])
        return dfs(0)

# dp
class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        dp = [False]*len(s)
        for i in range(len(s)):
            for word in wordDict:
                if i-len(word)+1 >= 0 and s[i-len(word)+1:i+1] == word and (i-len(word) < 0 or dp[i-len(word)]):
                    dp[i] = True
        return dp[-1]