谢谢小哥哥 | Merge pull request #271 from royIdoodle/master

0146. LRU Cache

docs/Leetcode_Solutions/JavaScript/0146._LRU_Cache.md

+# 0146. LRU Cache
+
+**<font color=red>难度: Medium</font>**
+
+## 刷题内容
+
+> 原题链接
+
+* https://leetcode.com/problems/lru-cache/
+
+> 内容描述
+
+Design and implement a data structure for [Least Recently Used (LRU) cache](https://en.wikipedia.org/wiki/Cache_replacement_policies#LRU). It should support the following operations: `get` and `put`.
+
+`get(key)` - Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1.
+`put(key, value)` - Set or insert the value if the key is not already present. When the cache reached its capacity, it should invalidate the least recently used item before inserting a new item.
+
+The cache is initialized with a **positive** capacity.
+
+**Follow up:**
+Could you do both operations in **O(1)** time complexity?
+
+**Example:**
+
+```
+LRUCache cache = new LRUCache( 2 /* capacity */ );
+
+cache.put(1, 1);
+cache.put(2, 2);
+cache.get(1);       // returns 1
+cache.put(3, 3);    // evicts key 2
+cache.get(2);       // returns -1 (not found)
+cache.put(4, 4);    // evicts key 1
+cache.get(1);       // returns -1 (not found)
+cache.get(3);       // returns 3
+cache.get(4);       // returns 4
+```
+
+ 
+
+## 解题方案
+
+> 思路1
+
+******- 时间复杂度: O(1)******- 空间复杂度: O(N)******
+
+这是一道数据结构的考题。
+
+简单的获取和插入，顺序不做考虑，所以数据结构用Object即可。
+
+但是重要的考点——[LRU](https://baike.baidu.com/item/LRU/1269842?fr=aladdin)，也就是删除最近没有使用的数据。所以想到用`队列`存在key列表：
+
+* 未超过存储上限时，每次`put`一个新数据时，向队列末尾插入当前`key`
+* 每次`get`时，如果key存在，则将对应的key从的队列中，移到对末尾
+* 在超过存储上限时，如进行`put`操作，则将队列首位删除掉
+
+> 执行用时 :492 ms, 在所有 JavaScript 提交中击败了35.29%的用户
+>
+> 内存消耗 :59.7 MB, 在所有 JavaScript 提交中击败了16.36%的用户
+
+```javascript
+/**
+ * @param {number} capacity
+ */
+var LRUCache = function(capacity) {
+  this.limit = capacity || 2
+  this.storage = {}
+  this.keyList = []
+};
+
+/**
+ * @param {number} key
+ * @return {number}
+ */
+LRUCache.prototype.get = function(key) {
+  if (this.storage.hasOwnProperty(key)) {
+    let index = this.keyList.findIndex(k => k === key)
+    this.keyList.splice(index, 1)
+    this.keyList.push(key)
+    return this.storage[key]
+  } else {
+    return -1
+  }
+};
+
+/**
+ * @param {number} key
+ * @param {number} value
+ * @return {void}
+ */
+LRUCache.prototype.put = function(key, value) {
+  // 判断容量
+  if (this.keyList.length >= this.limit && !this.storage.hasOwnProperty(key)) {
+    this.deleteLRU()
+  }
+  
+  // 存储数据
+  this.updateKeyList(key)
+  this.storage[key] = value
+};
+
+LRUCache.prototype.deleteLRU = function () {
+  delete this.storage[this.keyList.shift()]
+}
+
+LRUCache.prototype.updateKeyList = function (key) {
+  if (this.storage.hasOwnProperty(key)) {
+    var index = this.keyList.findIndex(k => key === k)
+    this.keyList.splice(index, 1)
+  }
+  this.keyList.push(key)
+}
+
+/**
+ * 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)
+ */
+
+
+```
+
+
+