https://leetcode-cn.com/problems/implement-trie-prefix-tree
实现一个 Trie (前缀树),包含 insert, search, 和 startsWith 这三个操作。
示例:
Trie trie = new Trie();
trie.insert("apple");
trie.search("apple"); // 返回 true
trie.search("app"); // 返回 false
trie.startsWith("app"); // 返回 true
trie.insert("app");
trie.search("app"); // 返回 true
说明:
你可以假设所有的输入都是由小写字母 a-z 构成的。
保证所有输入均为非空字符串。
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/implement-trie-prefix-tree
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
熟悉一下 Trie 的概念就可以了。
https://medium.com/basecs/trying-to-understand-tries-3ec6bede0014
Nodejs
const __main__ = function () {
const readline = require('readline')
const rl = readline.createInterface({
input: process.stdin,
output: process.stdout,
})
console.log('******输入******')
rl.prompt()
const lines = []
rl.on('line', line => lines.push(line))
rl.on('close', () => {
console.log('\n******输出******')
const test = (operations, params) => {
if (operations[0] === 'Trie') {
const trie = new Trie()
const output = [null]
for (let i = 1; i < operations.length; i++) {
const res = trie[operations[i]](...params[i])
output.push(res === void 0 ? null : res)
}
console.log(output)
} else {
console.log(Array(operations.length).fill(null))
}
}
while (lines.length >= 2) {
const params = lines.splice(0, 2)
test(...params.map(el => JSON.parse(el)))
}
})
}
- 时间复杂度:$O(L)$,L 是字符串长度,
insert
search
startsWith
操作都是。 - 空间复杂度:$O(M^{L})$,L 是字符串长度,M 是字符集中字符个数,如本题中 M 就是 26。
TypeScript Code
class TrieNode {
value: string
children: Array<TrieNode | null>
constructor(value) {
this.value = value
this.children = Array(26)
}
}
class Trie {
private root: TrieNode
constructor() {
this.root = this._getTrieNode('')
}
private _getTrieNode(value: string): TrieNode {
return new TrieNode(value)
}
private _char2Index(char: string): number {
return char.toLowerCase().charCodeAt(0) - 97
}
insert(word: string): void {
let crawl: TrieNode = this.root
for (let char of word) {
const index: number = this._char2Index(char)
if (!crawl.children[index]) {
crawl.children[index] = this._getTrieNode('')
}
crawl = crawl.children[index]
}
crawl.value = word
}
search(word: string): boolean {
let crawl: TrieNode = this.root
for (let char of word) {
const index: number = this._char2Index(char)
if (!crawl.children[index]) return false
crawl = crawl.children[index]
}
return crawl.value === word
}
startsWith(prefix: string): boolean {
let crawl: TrieNode = this.root
for (let char of prefix) {
const index: number = this._char2Index(char)
if (!crawl.children[index]) return false
crawl = crawl.children[index]
}
return true
}
}
/**
* Your Trie object will be instantiated and called as such:
* var obj = new Trie()
* obj.insert(word)
* var param_2 = obj.search(word)
* var param_3 = obj.startsWith(prefix)
*/
JavaScript Code
class TrieNode {
constructor(val) {
this.value = val
this.pointers = Array(26)
}
}
class Trie {
constructor() {
this.root = this._getTrieNode('')
}
/**
* @param {string} val
*/
_getTrieNode(val) {
return new TrieNode(val)
}
/**
* @param {string} char
* @returns {number}
*/
_char2Index(char) {
return char.toLowerCase().charCodeAt(0) - 97
}
/**
* Inserts a word into the trie.
* @param {string} word
* @return {void}
*/
insert(word) {
let crawl = this.root
for (let char of word) {
const index = this._char2Index(char)
if (!crawl.pointers[index]) {
crawl.pointers[index] = this._getTrieNode('')
}
crawl = crawl.pointers[index]
}
// Store the word in the last TrieNode as an end mark.
crawl.value = word
}
/**
* Returns if the word is in the trie.
* @param {string} word
* @return {boolean}
*/
search(word) {
let crawl = this.root
for (let char of word) {
const index = this._char2Index(char)
if (!crawl.pointers[index]) return false
crawl = crawl.pointers[index]
}
// If it has a stored value, it is the last TrieNode, i.e., the desired word is found.
// Otherwise, the word doesn't exist in Trie.
return !!crawl.value
}
/**
* Returns if there is any word in the trie that starts with the given prefix.
* @param {string} prefix
* @return {boolean}
*/
startsWith(prefix) {
let crawl = this.root
for (let char of prefix) {
const index = this._char2Index(char)
if (!crawl.pointers[index]) return false
crawl = crawl.pointers[index]
}
return true
}
}
官方题解
https://github.com/leetcode-pp/91alg-1/issues/68#issuecomment-657077878