Leetcode 139. Word Break

Seawolf

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

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，判定是否可以被空格拆分为一个或多个在字典中出现的单词。

说明：

拆分时可以重复使用字典中的单词。
你可以假设字典中没有重复的单词。
示例 1：

输入: s = "leetcode", wordDict = ["leet", "code"]
输出: true
解释: 返回 true 因为 "leetcode" 可以被拆分成 "leet code"。
示例 2：

输入: s = "applepenapple", wordDict = ["apple", "pen"]
输出: true
解释: 返回 true 因为 "applepenapple" 可以被拆分成 "apple pen apple"。
注意你可以重复使用字典中的单词。
示例 3：

输入: s = "catsandog", wordDict = ["cats", "dog", "sand", "and", "cat"]
输出: false

class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        dic = set(wordDict)
        hashmap = {}
        return self.helper(s, dic, hashmap)
    
    def helper(self, s: str, dic: List[str], hashmap) -> bool:
        if s in hashmap:
            return hashmap[s]
        if s in dic:
            hashmap[s] = True
            return True
        for i in range(1, len(s)):
            left = s[:i]
            right = s[i:]
            if self.helper(left, dic, hashmap) and right in dic:
                hashmap[s] = True
                return True
        hashmap[s] = False
        return False