Leetcode 139. Word Break

Seawolf

1. 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.
   
2. 
   
3. Note:
   
4. 
   
5. The same word in the dictionary may be reused multiple times in the segmentation.
   
6. You may assume the dictionary does not contain duplicate words.
   
7. Example 1:
   
8. 
   
9. Input: s = "leetcode", wordDict = ["leet", "code"]
   
10. Output: true
    
11. Explanation: Return true because "leetcode" can be segmented as "leet code".
    
12. Example 2:
    
13. 
    
14. Input: s = "applepenapple", wordDict = ["apple", "pen"]
    
15. Output: true
    
16. Explanation: Return true because "applepenapple" can be segmented as "apple pen apple".
    
17.              Note that you are allowed to reuse a dictionary word.
    
18. Example 3:
    
19. 
    
20. Input: s = "catsandog", wordDict = ["cats", "dog", "sand", "and", "cat"]
    
21. Output: false

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1. 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。
   
2. 
   
3. 说明：
   
4. 
   
5. 拆分时可以重复使用字典中的单词。
   
6. 你可以假设字典中没有重复的单词。
   
7. 示例 1：
   
8. 
   
9. 输入: s = "leetcode", wordDict = ["leet", "code"]
   
10. 输出: true
    
11. 解释: 返回 true 因为 "leetcode" 可以被拆分成 "leet code"。
    
12. 示例 2：
    
13. 
    
14. 输入: s = "applepenapple", wordDict = ["apple", "pen"]
    
15. 输出: true
    
16. 解释: 返回 true 因为 "applepenapple" 可以被拆分成 "apple pen apple"。
    
17. 注意你可以重复使用字典中的单词。
    
18. 示例 3：
    
19. 
    
20. 输入: s = "catsandog", wordDict = ["cats", "dog", "sand", "and", "cat"]
    
21. 输出: false

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1. class Solution:
   
2.     def wordBreak(self, s: str, wordDict: List[str]) -> bool:
   
3.         dic = set(wordDict)
   
4.         hashmap = {}
   
5.         return self.helper(s, dic, hashmap)
   
6.    
   
7.     def helper(self, s: str, dic: List[str], hashmap) -> bool:
   
8.         if s in hashmap:
   
9.             return hashmap[s]
   
10.         if s in dic:
    
11.             hashmap[s] = True
    
12.             return True
    
13.         for i in range(1, len(s)):
    
14.             left = s[:i]
    
15.             right = s[i:]
    
16.             if self.helper(left, dic, hashmap) and right in dic:
    
17.                 hashmap[s] = True
    
18.                 return True
    
19.         hashmap[s] = False
    
20.         return False

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