LeetCode 472. Concatenated Words

原题链接在这里：https://leetcode.com/problems/concatenated-words/

题目：

Given a list of words (without duplicates), please write a program that returns all concatenated words in the given list of words.

A concatenated word is defined as a string that is comprised entirely of at least two shorter words in the given array.

Example:

Input: ["cat","cats","catsdogcats","dog","dogcatsdog","hippopotamuses","rat","ratcatdogcat"]

Output: ["catsdogcats","dogcatsdog","ratcatdogcat"]

Explanation: "catsdogcats" can be concatenated by "cats", "dog" and "cats"; 
 "dogcatsdog" can be concatenated by "dog", "cats" and "dog"; 
"ratcatdogcat" can be concatenated by "rat", "cat", "dog" and "cat".

Note:

1. The number of elements of the given array will not exceed 10,000
2. The length sum of elements in the given array will not exceed 600,000.
3. All the input string will only include lower case letters.
4. The returned elements order does not matter.

题解：

For each word, it could be composed by shorter words.

Sort the words with length, then for each word check if it could be composed by previous words.

Time Complexity: O(m*logm + m*n^2). m = words.length. n = longest word length.

Space: O(m). 

AC Java:

class Solution {
    public List<String> findAllConcatenatedWordsInADict(String[] words) {
        List<String> res = new ArrayList<>();
        if(words == null || words.length == 0){
            return res;
        }
        
        Arrays.sort(words, (a, b) -> a.length() - b.length());
        HashSet<String> hs = new HashSet<>();
        
        for(int i = 1; i < words.length; i++){
            hs.add(words[i - 1]);
            if(couldCompose(words[i], hs)){
                res.add(words[i]);
            }
        }
        
        return res;
    }
    
    private boolean couldCompose(String s, Set<String> hs){
        if(s == null || s.length() == 0 || hs == null || hs.size() == 0){
            return false;
        }
        
        int n = s.length();
        boolean [] dp = new boolean[n + 1];
        dp[0] = true;
        for(int i = 1; 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];
    }
    
}

类似Word Break.