LeetCode 139. Word Break

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

题目：

Given a string s and a dictionary of strings wordDict, return true if s can be segmented into a space-separated sequence of one or more dictionary words.

Note that the same word in the dictionary may be reused multiple times in the segmentation.

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

Constraints:

• 1 <= s.length <= 300
• 1 <= wordDict.length <= 1000
• 1 <= wordDict[i].length <= 20
• s and wordDict[i] consist of only lowercase English letters.
• All the strings of wordDict are unique.

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:

class Solution {
    public boolean wordBreak(String s, List<String> wordDict) {
        if(s == null || s.length() == 0){
            return true;
        }
        
        Set<String> hs = new HashSet<>(wordDict);
        int n = s.length();
        boolean [] dp = new boolean[n + 1];
        dp[0] = true;
        for(int i = 0; i <= n; i++){
            for(int j = 0; j < i; j++){
                if(dp[j] && hs.contains(s.substring(j, i))){
                    dp[i] = true;
                    break;
                }
            }            
        }

        return dp[n];
    }
}

AC Python:

class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        dp = [False] * (len(s) + 1)
        dp[0] = True
        for i in range(len(s) + 1):
            for j in range(i):
                if dp[j] and s[j: i] in wordDict:
                    dp[i] = True
                    continue
        
        return dp[-1]

跟上Word Break II.

类似Concatenated Words.