lru

Author: luzhipeng

Diff for: 146.lru-cache.md

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+## 题目地址
+https://leetcode.com/problems/lru-cache/description/
+
+## 题目描述
+
+```
+Design and implement a data structure for Least Recently Used (LRU) cache. 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.
+
+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
+
+```
+
+## 思路
+
+由于是保留是最近使用的N条数据，这就和队列的特性很符合， 先进入队列的，先出队列。
+
+因此思路就是用一个队列来记录目前缓存的所有key， 每次push都进行判断，如果
+超出最大容量限制则进行清除缓存的操作， 具体清除谁就按照刚才说的队列方式进行处理，同时对key进行入队操作。
+
+get的时候，如果缓存中有，则调整队列（具体操作为删除指定元素和入队两个操作）。 缓存中没有则返回-1
+
+## 关键点解析
+
+- 队列简化操作
+
+- 队列的操作是这道题的灵魂， 很容易少考虑情况
+
+
+## 代码
+
+```js
+/*
+ * @lc app=leetcode id=146 lang=javascript
+ *
+ * [146] LRU Cache
+ *
+ * https://leetcode.com/problems/lru-cache/description/
+ *
+ * algorithms
+ * Hard (24.17%)
+ * Total Accepted:    272.8K
+ * Total Submissions: 1.1M
+ * Testcase Example:  '["LRUCache","put","put","get","put","get","put","get","get","get"]\n[[2],[1,1],[2,2],[1],[3,3],[2],[4,4],[1],[3],[4]]'
+ *
+ * 
+ * Design and implement a data structure for Least Recently Used (LRU) cache.
+ * 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.
+ * 
+ * 
+ * Follow up:
+ * Could you do both operations in O(1) time complexity?
+ * 
+ * Example:
+ * 
+ * LRUCache cache = new LRUCache( 2 );
+ * 
+ * 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
+ * 
+ * 
+ */
+/**
+ * @param {number} capacity
+ */
+var LRUCache = function(capacity) {
+    this.cache = {};
+    this.capacity = capacity;
+    this.size = 0;
+    this.queue = [];
+};
+
+/** 
+ * @param {number} key
+ * @return {number}
+ */
+LRUCache.prototype.get = function(key) {
+    const hit = this.cache[key];
+
+    if (hit !== undefined) {
+        this.queue = this.queue.filter(q => q !== key);
+        this.queue.push(key);
+        return hit;
+    }
+    return -1;
+};
+
+/** 
+ * @param {number} key 
+ * @param {number} value
+ * @return {void}
+ */
+LRUCache.prototype.put = function(key, value) {
+    const hit = this.cache[key];
+
+    // update cache
+    this.cache[key] = value;
+
+    if (!hit) {
+        // invalid cache and resize size;
+        if (this.size === this.capacity) {
+            // invalid cache
+            const key = this.queue.shift();
+            this.cache[key] = undefined;
+        } else {
+            this.size = this.size + 1;
+        }
+        this.queue.push(key);
+    } else {
+        this.queue = this.queue.filter(q => q !== key);
+        this.queue.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)
+ */
+
+
+```