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Add solution 146、460
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YDZ
71
leetcode/0146.LRU-Cache/146. LRU Cache.go
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71
leetcode/0146.LRU-Cache/146. LRU Cache.go
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package leetcode
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type LRUCache struct {
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head, tail *Node
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Keys map[int]*Node
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Cap int
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}
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type Node struct {
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Key, Val int
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Prev, Next *Node
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}
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func Constructor(capacity int) LRUCache {
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return LRUCache{Keys: make(map[int]*Node), Cap: capacity}
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}
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func (this *LRUCache) Get(key int) int {
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if node, ok := this.Keys[key]; ok {
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this.Remove(node)
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this.Add(node)
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return node.Val
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}
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return -1
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}
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func (this *LRUCache) Put(key int, value int) {
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if node, ok := this.Keys[key]; ok {
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node.Val = value
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this.Remove(node)
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this.Add(node)
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return
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} else {
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node = &Node{Key: key, Val: value}
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this.Keys[key] = node
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this.Add(node)
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}
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if len(this.Keys) > this.Cap {
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delete(this.Keys, this.tail.Key)
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this.Remove(this.tail)
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}
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}
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func (this *LRUCache) Add(node *Node) {
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node.Prev = nil
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node.Next = this.head
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if this.head != nil {
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this.head.Prev = node
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}
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this.head = node
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if this.tail == nil {
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this.tail = node
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this.tail.Next = nil
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}
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}
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func (this *LRUCache) Remove(node *Node) {
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if node == this.head {
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this.head = node.Next
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node.Next = nil
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return
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}
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if node == this.tail {
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this.tail = node.Prev
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node.Prev.Next = nil
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node.Prev = nil
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return
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}
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node.Prev.Next = node.Next
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node.Next.Prev = node.Prev
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}
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40
leetcode/0146.LRU-Cache/146. LRU Cache_test.go
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40
leetcode/0146.LRU-Cache/146. LRU Cache_test.go
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package leetcode
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import (
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"fmt"
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"testing"
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)
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func Test_Problem147(t *testing.T) {
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obj := Constructor(2)
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fmt.Printf("obj = %v\n", MList2Ints(&obj))
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obj.Put(1, 1)
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fmt.Printf("obj = %v\n", MList2Ints(&obj))
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obj.Put(2, 2)
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fmt.Printf("obj = %v\n", MList2Ints(&obj))
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param1 := obj.Get(1)
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fmt.Printf("param_1 = %v obj = %v\n", param1, MList2Ints(&obj))
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obj.Put(3, 3)
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fmt.Printf("obj = %v\n", MList2Ints(&obj))
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param1 = obj.Get(2)
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fmt.Printf("param_1 = %v obj = %v\n", param1, MList2Ints(&obj))
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obj.Put(4, 4)
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fmt.Printf("obj = %v\n", MList2Ints(&obj))
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param1 = obj.Get(1)
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fmt.Printf("param_1 = %v obj = %v\n", param1, MList2Ints(&obj))
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param1 = obj.Get(3)
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fmt.Printf("param_1 = %v obj = %v\n", param1, MList2Ints(&obj))
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param1 = obj.Get(4)
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fmt.Printf("param_1 = %v obj = %v\n", param1, MList2Ints(&obj))
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}
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func MList2Ints(lru *LRUCache) [][]int {
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res := [][]int{}
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head := lru.head
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for head != nil {
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tmp := []int{head.Key, head.Val}
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res = append(res, tmp)
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head = head.Next
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}
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return res
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}
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134
leetcode/0146.LRU-Cache/README.md
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134
leetcode/0146.LRU-Cache/README.md
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# [146. LRU Cache](https://leetcode.com/problems/lru-cache/)
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## 题目
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Design a data structure that follows the constraints of a **[Least Recently Used (LRU) cache](https://en.wikipedia.org/wiki/Cache_replacement_policies#LRU)**.
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Implement the `LRUCache` class:
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- `LRUCache(int capacity)` Initialize the LRU cache with **positive** size `capacity`.
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- `int get(int key)` Return the value of the `key` if the key exists, otherwise return `1`.
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- `void put(int key, int value)` Update the value of the `key` if the `key` exists. Otherwise, add the `key-value` pair to the cache. If the number of keys exceeds the `capacity` from this operation, **evict** the least recently used key.
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**Follow up:**Could you do `get` and `put` in `O(1)` time complexity?
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**Example 1:**
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```
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Input
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["LRUCache", "put", "put", "get", "put", "get", "put", "get", "get", "get"]
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[[2], [1, 1], [2, 2], [1], [3, 3], [2], [4, 4], [1], [3], [4]]
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Output
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[null, null, null, 1, null, -1, null, -1, 3, 4]
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Explanation
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LRUCache lRUCache = new LRUCache(2);
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lRUCache.put(1, 1); // cache is {1=1}
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lRUCache.put(2, 2); // cache is {1=1, 2=2}
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lRUCache.get(1); // return 1
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lRUCache.put(3, 3); // LRU key was 2, evicts key 2, cache is {1=1, 3=3}
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lRUCache.get(2); // returns -1 (not found)
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lRUCache.put(4, 4); // LRU key was 1, evicts key 1, cache is {4=4, 3=3}
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lRUCache.get(1); // return -1 (not found)
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lRUCache.get(3); // return 3
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lRUCache.get(4); // return 4
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```
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**Constraints:**
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- `1 <= capacity <= 3000`
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- `0 <= key <= 3000`
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- `0 <= value <= 104`
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- At most `3 * 104` calls will be made to `get` and `put`.
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## 题目大意
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运用你所掌握的数据结构，设计和实现一个 LRU (最近最少使用) 缓存机制 。
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实现 LRUCache 类：
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- LRUCache(int capacity) 以正整数作为容量 capacity 初始化 LRU 缓存
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- int get(int key) 如果关键字 key 存在于缓存中，则返回关键字的值，否则返回 -1 。
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- void put(int key, int value) 如果关键字已经存在，则变更其数据值；如果关键字不存在，则插入该组「关键字-值」。当缓存容量达到上限时，它应该在写入新数据之前删除最久未使用的数据值，从而为新的数据值留出空间。
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进阶：你是否可以在 O(1) 时间复杂度内完成这两种操作？
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## 解题思路
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- 这一题是 LRU 经典面试题，详细解释见第三章模板。
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## 代码
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```go
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package leetcode
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type LRUCache struct {
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head, tail *Node
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Keys map[int]*Node
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Cap int
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}
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type Node struct {
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Key, Val int
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Prev, Next *Node
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}
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func Constructor(capacity int) LRUCache {
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return LRUCache{Keys: make(map[int]*Node), Cap: capacity}
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}
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func (this *LRUCache) Get(key int) int {
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if node, ok := this.Keys[key]; ok {
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this.Remove(node)
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this.Add(node)
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return node.Val
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}
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return -1
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}
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func (this *LRUCache) Put(key int, value int) {
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if node, ok := this.Keys[key]; ok {
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node.Val = value
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this.Remove(node)
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this.Add(node)
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return
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} else {
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node = &Node{Key: key, Val: value}
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this.Keys[key] = node
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this.Add(node)
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}
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if len(this.Keys) > this.Cap {
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delete(this.Keys, this.tail.Key)
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this.Remove(this.tail)
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}
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}
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func (this *LRUCache) Add(node *Node) {
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node.Prev = nil
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node.Next = this.head
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if this.head != nil {
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this.head.Prev = node
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}
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this.head = node
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if this.tail == nil {
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this.tail = node
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this.tail.Next = nil
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}
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}
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func (this *LRUCache) Remove(node *Node) {
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if node == this.head {
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this.head = node.Next
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node.Next = nil
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return
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}
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if node == this.tail {
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this.tail = node.Prev
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node.Prev.Next = nil
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node.Prev = nil
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return
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}
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node.Prev.Next = node.Next
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node.Next.Prev = node.Prev
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}
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```
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