js iterator protocol & next() All In One
<h1 id="js-iterator-protocol--next-all-in-one">js iterator protocol & next() All In One</h1><blockquote>
<p><code>.next()</code></p>
</blockquote>
<pre><code class="language-js">// Satisfies both the Iterator Protocol and Iterable
const myIterator = {
next: function() {
// ...
},
: function() { return this; }
};
</code></pre>
<p><img src="https://img2022.cnblogs.com/blog/740516/202206/740516-20220607115549439-1802628900.png" alt="" loading="lazy"></p>
<h2 id="iterator-protocol">iterator protocol</h2>
<blockquote>
<p>迭代器协议</p>
</blockquote>
<pre><code class="language-js">// .next()
// .next().value
// .next().done
</code></pre>
<p>https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Iteration_protocols#the_iterator_protocol</p>
<h2 id="iterators--generators">iterators & generators</h2>
<blockquote>
<p>迭代器、生成器</p>
</blockquote>
<pre><code class="language-js">
</code></pre>
<pre><code class="language-js">
</code></pre>
<p>https://developer.mozilla.org/en-US/docs/Web/JavaScript/Guide/Iterators_and_Generators</p>
<h2 id="demo">demo</h2>
<pre><code class="language-js">
"use strict";
/**
*
* @author xgqfrms
* @license MIT
* @copyright xgqfrms
* @created 2020-11-26
* @modified
*
* @description 146. LRU Cache
* @description 146. LRU Cache
* @difficulty Medium
* @complexity O(n)
* @time O(n)
* @augments
* @example
* @link https://leetcode.com/problems/lru-cache/
* @link https://leetcode-cn.com/problems/lru-cache/
* @link https://en.wikipedia.org/wiki/Cache_replacement_policies#LRU 最近最少使用（LRU）缓存算法
* @solutions
*
* @best_solutions
*
*/
const log = console.log;
/**
* @param {number} capacity
*/
var LRUCache = function(capacity) {
this.capacity = capacity;
this.map = new Map();
};
/**
* @param {number} key
* @return {number}
*/
LRUCache.prototype.get = function(key) {
// ✅ 优先处理代码少的逻辑，错误优先
if(!this.map.has(key)) {
return -1;
} else {
const value = this.map.get(key)
// 读取后，更新顺序
this.put(key, value);
return value;
}
// if(this.map.has(key)) {
// const value = this.map.get(key)
// // 读取后，更新顺序
// this.put(key, value);
// return value;
// } else {
// return -1;
// }
};
/**
* @param {number} key
* @param {number} value
* @return {void}
*/
LRUCache.prototype.put = function(key, value) {
// 替换式更新，先删除
if(this.map.has(key)) {
this.map.delete(key);
}
// 达到容量限制，先删除最近最少使用的 map 第一个元素
if(this.map.size === this.capacity) {
const oldKey = this.map.keys().next().value;
this.map.delete(oldKey);
}
// 再重新添加
this.map.set(key, value);
};
/**
* Your LRUCache object will be instantiated and called as such:
* var obj = new LRUCache(capacity)
* var param_1 = obj.get(key)
* obj.put(key,value)
*/
/*
lRUCache = new LRUCache(2);
lRUCache.put(1, 1);
// cache is {1=1}
lRUCache.put(2, 2);
// cache is {1=1, 2=2}
lRUCache.get(1);
// return 1
lRUCache.put(3, 3);
// LRU key was 2, evicts key 2, cache is {1=1, 3=3}
lRUCache.get(2);
// returns -1 (not found)
lRUCache.put(4, 4);
// LRU key was 1, evicts key 1, cache is {4=4, 3=3}
lRUCache.get(1);
// return -1 (not found)
lRUCache.get(3);
// return 3
lRUCache.get(4);
// return 4
*/
</code></pre>
<p><img src="https://img2022.cnblogs.com/blog/740516/202206/740516-20220607122343942-529774300.png" alt="" loading="lazy"></p>
<p>https://leetcode.com/problems/lru-cache/</p>
<p>https://leetcode.cn/problems/lru-cache/</p>
<h2 id="refs">refs</h2>
<p>https://www.cnblogs.com/xgqfrms/p/16344654.html</p>
<p>https://leetcode.com/problems/lru-cache/discuss/2074758/Simple-Javascript-solution-or-Easy-to-understand</p>
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</h3></blockquote>
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<p>本文首发于博客园，作者：xgqfrms，原文链接：https://www.cnblogs.com/xgqfrms/p/16351282.html</p>
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来源：https://www.cnblogs.com/xgqfrms/p/16351282.html
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