From 3770b30040ded9199dd5c948e3379a0efce9774b Mon Sep 17 00:00:00 2001
From: yige <a@wyr.me>
Date: Mon, 25 May 2020 22:27:39 +0800
Subject: [PATCH] =?UTF-8?q?add:=20LRU=E7=BC=93=E5=AD=98=E6=9C=BA=E5=88=B6?=
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---
 README.md                    |  5 +++++
 src/array/lru-cache.js       | 21 +++++++++++++++++++++
 test/array/lru-cache.test.js | 15 +++++++++++++++
 3 files changed, 41 insertions(+)
 create mode 100644 src/array/lru-cache.js
 create mode 100644 test/array/lru-cache.test.js

diff --git a/README.md b/README.md
index 3a086b2..ba7864c 100644
--- a/README.md
+++ b/README.md
@@ -343,6 +343,11 @@ LeetCode 与 LintCode 解题记录。此为个人练习仓库，代码中对重
   - LeetCode 860. 柠檬水找零 <https://leetcode-cn.com/problems/lemonade-change/>
   - LintCode 1509. 柠檬水找零 <https://www.lintcode.com/problem/lemonade-change/description>
 
+- [LRU缓存机制](src/array/lru-cache.js)
+
+  - LeetCode 146. LRU缓存机制 <https://leetcode-cn.com/problems/lru-cache/>
+  - LintCode 134. LRU缓存策略 <https://www.lintcode.com/problem/lru-cache/description>
+
 ## 栈
 
 - [最大矩阵](src/stack/maximal-rectangle.js)
diff --git a/src/array/lru-cache.js b/src/array/lru-cache.js
new file mode 100644
index 0000000..4c5d1d0
--- /dev/null
+++ b/src/array/lru-cache.js
@@ -0,0 +1,21 @@
+export default class LRUCache {
+  constructor (capacity) {
+    this.max = capacity; this.map = new Map()
+  }
+
+  get (key) {
+    const value = this.map.get(key)
+    if (value !== undefined) {
+      this.map.delete(key)
+      this.map.set(key, value)
+      return value
+    }
+    return -1
+  }
+
+  put (key, value) {
+    this.map.delete(key)
+    this.map.set(key, value)
+    if (this.map.size > this.max) this.map.delete(this.map.keys().next().value)
+  }
+}
diff --git a/test/array/lru-cache.test.js b/test/array/lru-cache.test.js
new file mode 100644
index 0000000..bed276e
--- /dev/null
+++ b/test/array/lru-cache.test.js
@@ -0,0 +1,15 @@
+import LRUCache from '../../src/array/lru-cache'
+
+test('LRU缓存机制', () => {
+  const cache = new LRUCache(2 /* 缓存容量 */)
+
+  cache.put(1, 1)
+  cache.put(2, 2)
+  expect(cache.get(1)).toBe(1) // 返回  1
+  cache.put(3, 3) // 该操作会使得密钥 2 作废
+  expect(cache.get(2)).toBe(-1) // 返回 -1 (未找到)
+  cache.put(4, 4) // 该操作会使得密钥 1 作废
+  expect(cache.get(1)).toBe(-1) // 返回 -1 (未找到)
+  expect(cache.get(3)).toBe(3) // 返回  3
+  expect(cache.get(4)).toBe(4) // 返回  4
+})