#yyds干货盘点# LeetCode程序员面试金典：连接词

题目

给你一个 不含重复 单词的字符串数组 words ，请你找出并返回 words 中的所有 连接词 。

连接词 定义为：一个完全由给定数组中的至少两个较短单词（不一定是不同的两个单词）组成的字符串。

示例 1：

输入：words = ["cat","cats","catsdogcats","dog","dogcatsdog","hippopotamuses","rat","ratcatdogcat"]

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

解释："catsdogcats" 由 "cats", "dog" 和 "cats" 组成;  

    "dogcatsdog" 由 "dog", "cats" 和 "dog" 组成;  

    "ratcatdogcat" 由 "rat", "cat", "dog" 和 "cat" 组成。

示例 2：

输入：words = ["cat","dog","catdog"]

输出：["catdog"]

代码实现

class Solution {
    Trie trie = new Trie();

    public List<String> findAllConcatenatedWordsInADict(String[] words) {
        List<String> ans = new ArrayList<String>();
        Arrays.sort(words, (a, b) -> a.length() - b.length());
        for (int i = 0; i < words.length; i++) {
            String word = words[i];
            if (word.length() == 0) {
                continue;
            }
            boolean[] visited = new boolean[word.length()];
            if (dfs(word, 0, visited)) {
                ans.add(word);
            } else {
                insert(word);
            }
        }
        return ans;
    }

    public boolean dfs(String word, int start, boolean[] visited) {
        if (word.length() == start) {
            return true;
        }
        if (visited[start]) {
            return false;
        }
        visited[start] = true;
        Trie node = trie;
        for (int i = start; i < word.length(); i++) {
            char ch = word.charAt(i);
            int index = ch - 'a';
            node = node.children[index];
            if (node == null) {
                return false;
            }
            if (node.isEnd) {
                if (dfs(word, i + 1, visited)) {
                    return true;
                }
            }
        }
        return false;
    }
    
    public void insert(String word) {
        Trie node = trie;
        for (int i = 0; i < word.length(); i++) {
            char ch = word.charAt(i);
            int index = ch - 'a';
            if (node.children[index] == null) {
                node.children[index] = new Trie();
            }
            node = node.children[index];
        }
        node.isEnd = true;
    }
}

class Trie {
    Trie[] children;
    boolean isEnd;

    public Trie() {
        children = new Trie[26];
        isEnd = false;
    }
}