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:

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

这题后面还有一个进阶版本的Word Break II.

类似Concatenated Words.