LeetCode 139. Word Break

原题链接在这里：https://leetcode.com/problems/word-break/

题目：

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

题解：

Let dp[i] denotes up to index i, s.substring(0,i) could break into words or not.

For all the j from 0 to i, if dp[j] is true, and s.substring(j,i) is in the dictionary, then dp[i] is true.

Time Complexity: O(s.length() ^ 3). since after Java 7, substring take O(n).

Space: O(s.length()).

AC Java:

 1 class Solution {
 2     public boolean wordBreak(String s, List<String> wordDict) {
 3         if(s == null || s.length() == 0){
 4             return true;
 5         }
 6         
 7         HashSet<String> hs = new HashSet<String>(wordDict);
 8         boolean [] dp = new boolean[s.length()+1];
 9         dp[0] = true;
10         for(int i = 0; i<=s.length(); i++){
11             for(int j = 0; j<i; j++){
12                 if(dp[j] && hs.contains(s.substring(j, i))){
13                     dp[i] = true;
14                     continue;
15                 }
16             }
17         }
18         
19         return dp[s.length()];
20     }
21 }

AC Python:

 1 class Solution:
 2     def wordBreak(self, s: str, wordDict: List[str]) -> bool:
 3         dp = [False] * (len(s) + 1)
 4         dp[0] = True
 5         for i in range(len(s) + 1):
 6             for j in range(i):
 7                 if dp[j] and s[j: i] in wordDict:
 8                     dp[i] = True
 9                     continue
10         
11         return dp[-1]

跟上Word Break II.

类似Concatenated Words.