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1.描述
720. 词典中最长的单词 - 力扣(LeetCode) (leetcode-cn.com)
给出一个字符串数组 words
组成的一本英语词典。返回 words
中最长的一个单词,该单词是由 words
词典中其他单词逐步添加一个字母组成。
若其中有多个可行的答案,则返回答案中字典序最小的单词。若无答案,则返回空字符串。
示例 1:
输入:words = ["w","wo","wor","worl", "world"]
输出:"world"
解释: 单词"world"可由"w", "wo", "wor", 和 "worl"逐步添加一个字母组成。
示例 2:
输入:words = ["a", "banana", "app", "appl", "ap", "apply", "apple"]
输出:"apple"
解释:"apply" 和 "apple" 都能由词典中的单词组成。但是 "apple" 的字典序小于 "apply"
提示:
- 1 <= words.length <= 1000
- 1 <= words[i].length <= 30
- 所有输入的字符串 words[i] 都只包含小写字母。
2.分析
- 按字典顺序对words进行排序。从而按层构建trie tree。从而可以在构建阶段就确定哪些字符串的前缀都是words中包含的字符串。
- 逐个遍历words,并进行插入。
- 全部插入完成以后,直接从字典树对象longestwords上获取首位字符串,就是结果。(因为输入已经按字典顺序排序了。前面的是字典顺序小的)
3.AC代码
function longestWord(words: string[]): string {
let trieTree = new Trie();
//为了正确的维护trie的不同节点上的终结状态 需要按字符串长度顺序升序构建trie tree.
words.sort((a, b) => {
return a.localeCompare(b);
});
for (let i = 0; i < words.length; i++) {
const curWord = words[i];
trieTree.insert(curWord);
}
if (trieTree.longestwords.length > 0){
return trieTree.longestwords[0];
}
return "";
};
class Trie {
private root: TrieNode;
public longestwords: string[] = [];
constructor(){
this.root = new TrieNode("/", 0);
}
//往trie树中插入一个字符串
insert(word: string){
let p:TrieNode = this.root;
let isEveryCharEndingChar = true;
for (let i = 0; i < word.length; i++) {
const char = word[i];
if (!p.children.has(char)){
let newNode = new TrieNode(char, p.level + 1);
p.children.set(char, newNode);
}
//@ts-ignore
p = p.children.get(char);
if (i === word.length - 1){
p.isEndingChar = true;
}
isEveryCharEndingChar = isEveryCharEndingChar && p.isEndingChar;
}
if (isEveryCharEndingChar){
if (this.longestwords.length === 0){
this.longestwords.push(word);
} else if (this.longestwords[0].length < word.length){
this.longestwords = [word];
} else if (this.longestwords[0].length === word.length){
this.longestwords.push(word);
}
}
}
}
class TrieNode {
data: string = ""
level: number = -1
isEndingChar: boolean = false
children: Map<string, TrieNode>;
constructor(data: string, level: number){
this.data = data;
this.level = level;
this.children = new Map();
}
}
4.总结
前缀树免DFS,直接构建过程求解最长单词即可