字典中的最長單詞 Longest Word in Dictionary

問題：

Given a list of strings words representing an English Dictionary, find the longest word in words that can be built one character at a time by other words in words. If there is more than one possible answer, return the longest word with the smallest lexicographical order.

If there is no answer, return the empty string.

Example 1:

Input: 
words = ["w","wo","wor","worl", "world"]
Output: "world"
Explanation: 
The word "world" can be built one character at a time by "w", "wo", "wor", and "worl".

Example 2:

Input: 
words = ["a", "banana", "app", "appl", "ap", "apply", "apple"]
Output: "apple"
Explanation: 
Both "apply" and "apple" can be built from other words in the dictionary. However, "apple" is lexicographically smaller than "apply".

Note:

• All the strings in the input will only contain lowercase letters.
• The length of words will be in the range [1, 1000].
• The length of words[i] will be in the range [1, 30].

解決：

【題意】給定一個字典，是個字符串數組，從單個字符開始拼，返回最長能組成單詞，注意中間生成的字符串也要在字典中存在，並且當組成的單詞長度相等時，返回字母順序小的那個。

① 先將單詞按照字典序排序，爲了快速的查找某個單詞是否在字典中存在，將字典中的單詞存儲到set集合中。

class Solution { //42ms
    public String longestWord(String[] words) {
        Arrays.sort(words);//按照字典序排序
        Set<String> set = new HashSet<>();//用於存儲字典中已經存在的字符
        String res = "";
        for (String word : words){
            if (word.length() == 1 || set.contains(word.substring(0,word.length() - 1))){
                res = word.length() > res.length() ? word : res;
                set.add(word);
            }
        }
        return res;
    }
}

② 使用前綴樹，bfs查找。

class Solution { //20ms
    class TrieNode{
        TrieNode[] children;
        boolean isWord;
        String word;
        public TrieNode(){
            children = new TrieNode[26];
        }
    }
    class Trie{
        private TrieNode root;
        public Trie(){
            root = new TrieNode();
        }
        public void insert(String word){
            TrieNode node = root;
            for (int i = 0;i < word.length();i ++){
                int index = word.charAt(i) - 'a';
                if (node.children[index] == null){
                    node.children[index] = new TrieNode();
                }
                node = node.children[index];
            }
            node.isWord = true;
            node.word = word;
        }
        public String findLongestWord(){
            String res = null;
            Queue<TrieNode> queue = new LinkedList<>();
            queue.offer(root);
            while(! queue.isEmpty()){
                int count = queue.size();
                for (int i = 0;i < count;i ++){
                    TrieNode node = queue.poll();
                    for (int j = 25;j >= 0;j --){
                        if (node.children[j] != null && node.children[j].isWord){
                            res = node.children[j].word;
                            queue.offer(node.children[j]);
                        }
                    }
                }
            }
            return res;
        }
    }
    public String longestWord(String[] words) {
        Trie trie = new Trie();
        for (String word : words){
            trie.insert(word);
        }
        return trie.findLongestWord();
    }
}

③ 前綴樹+dfs查找

class Solution { //18ms     class TrieNode{        TrieNode[] children;        String word;        public TrieNode(){            children = new TrieNode[26];        }     }     TrieNode root = new TrieNode();     public String longestWord(String[] words){         root.word = "";         for (String word : words){             insert(word);         }         TrieNode cur = root;         return dfs(cur);     }     public void insert(String word){         TrieNode cur = root;         for (char c : word.toCharArray()){             int index = c - 'a';             if (cur.children[index] == null){                 cur.children[index] = new TrieNode();             }             cur = cur.children[index];         }         cur.word = word;     }     public String dfs(TrieNode cur){         String res = cur.word;         for (int i = 0;i < 26;i ++){             if (cur.children[i] != null && cur.children[i].word != null){                 String tmp = dfs(cur.children[i]);                 if (tmp.length() > res.length()){                     res = tmp;                 }             }         }         return res;     } }