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. class Solution {
   
2.     HashMap<String,Boolean> map = new HashMap<>();
   
3.     public boolean wordBreak(String s, List<String> wordDict) {
   
4.         return help(s, wordDict);
   
5.     }
   
6.    
   
7.     public boolean help(String s, List<String> Dict){
   
8.         if(map.containsKey(s)) return map.get(s);
   
9.         if(Dict.contains(s)) {
   
10.             map.put(s,true);
    
11.             return true;
    
12.         }
    
13.         
    
14.         for(int i = 1; i< s.length() ; i++){
    
15.             String left = s.substring(0,i);
    
16.             String right = s.substring(i);
    
17.             if(Dict.contains(right) && help(left,Dict)) {
    
18.                 map.put(s,true);
    
19.                 return true;
    
20.             }
    
21.         }
    
22.         
    
23.         map.put(s,false);
    
24.         return false;
    
25.     }
    
26. }

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