資訊專欄INFORMATION COLUMN
Problem
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].
class Solution {
class Trie {
Node root;
class Node {
boolean isWord;
Node[] children = new Node[26];
}
public Trie() {
root = new Node();
root.isWord = true;
}
void insert (String str) {
Node node = root;
for (char ch: str.toCharArray()) {
int index = ch - "a";
if (node.children[index] == null) {
node.children[index] = new Node();
}
node = node.children[index];
}
node.isWord = true;
}
boolean hasWord(String word) {
Node node = root;
for (char ch: word.toCharArray()) {
int index = ch - "a";
if (node.children[index] == null || !node.isWord) return false;
node = node.children[index];
}
return true;
}
}
public String longestWord(String[] words) {
Trie trie = new Trie();
for (String word: words) {
trie.insert(word);
}
String res = "";
for (String word: words) {
if (trie.hasWord(word) &&
(word.length() > res.length() || (word.length() == res.length() && word.compareTo(res) < 0))) {
res = word;
}
}
return res;
}
}
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