Leetcode-139. Word Break 单词拆分（DP）

题目

给定一个非空字符串 s 和一个包含非空单词列表的字典 wordDict，判定 s 是否可以被空格拆分为一个或多个在字典中出现的单词。
链接：https://leetcode.com/problems/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:

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.

思路及代码

• dp[i]： 字符串到第i位为止能不能实现wordbreak
• dp[i] = dp[i - len(word)] and word in word_dict

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

复杂度

T = $O(n^2)$
S = $O(n^2)$