Longest Word in Dictionary

问题：

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]

.

解决：

【题意】给定一个字典，是个字符串数组，从单个字符开始拼，返回最长能组成单词，注意中间生成的字符串也要在字典中存在，而且当组成的单词长度相等时，返回字母顺序小的那个。

① 先将单词按照字典序排序，为了快速的查找某个单词是否在字典中存在，将字典中的单词存储到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<>();//用于存储字典中已经存在的字符
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 使用前缀树，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 前缀树+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;
}
}