題目

Design and implement a data structure for Least Recently Used (LRU) cache. It should support the following operations:?get?and?set.

get(key)?- Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1.
set(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.

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思路

1. 雙向鏈表支持常數時間增刪元素, 單向鏈表配合 hash table 也可以實現, 但是實現相當復雜

2. hash table 存儲鏈表中的元素

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總結

1. 雙向鏈表要處理的情況比單向鏈表更復雜

2. 單向鏈表處理時, 僅需考慮 head 的問題, 但雙向鏈表則需要考慮

　　a) 增元素時, 考慮鏈表是否為空即 head 問題

　　b) 刪元素時, 要考慮被刪除的元素是否為 head?

　　c) LRU 操作, 前移元素時, 需要考慮被前移的元素是否為 Head, 這一個 bug 浪費了 2 個小時

　　d) 有一個優化, 即 LRU set miss 時, 不需要顯式的刪除再增加節點, 僅需將 tail 節點的 key, value 修改 然后 head 前進一步即可

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代碼

class Node { public:int key, value;Node *next, *pre;Node(int _key, int _value):key(_key), value(_value), next(NULL), pre(NULL){} }; class LRUCache{ public:int capacity;int curSize;unordered_map<int, Node*> mp;Node *head;LRUCache(int capacity) {this->capacity = capacity;curSize = 0;}int get(int key) {if(mp.find(key) == mp.end())return -1;int val = mp.find(key)->second->value;set(key, val);return val;}void set(int key, int value) {int ans = 0;if(mp.find(key) == mp.end())ans = -1;if(ans == -1) { // 新建一個 itemif(curSize >= capacity) { // 已滿, 需要刪除一個 itemif(capacity == 1) {mp.erase(mp.find(head->key));head->key = key;head->value = value;mp[key] = head;return;}else{ // 模擬刪除隊尾元素Node *tail = head->pre;mp.erase(mp.find(tail->key));tail->key = key;tail->value = value;head = tail;mp[key] = head;return;}}// 未滿, 添加元素if(curSize == 0) {head = new Node(key, value);head->pre = head;head->next = head;mp[key] = head;curSize++;return ;}else {Node *al = new Node(key, value);al->pre = head->pre;al->next = head;al->pre->next = al;head->pre = al;head = al;mp[key] = al;curSize++;return;}}else{if(head->key == key) {head->value = value;mp[key] = head;return;}//cout << "here" << endl;Node *mid = mp.find(key)->second;mid->next->pre = mid->pre;mid->pre->next = mid->next;mid->value = value;mid->next = head;mid->pre = head->pre;head->pre = mid;mid->pre->next = mid;head = mid;mp[key] = mid; // 修改 mp[key]return;}} };

　　

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Update?

#include <iostream> #include <map> using namespace std;class ListNode1 { public:ListNode1(int _key, int _val) {key = _key;val = _val;pre = next = NULL;}ListNode1 *pre;int key;int val;ListNode1 *next; };class LRUCache{ public:LRUCache(int capacity) {this->capacity = capacity;head = NULL;size = 0;}int get(int key) {if(mp.count(key) == 0)return -1;int value = mp[key]->val;set(key, value);return value;}void set(int key, int value) {if(capacity == 1) {mp.clear();head = new ListNode1(key, value);mp[key] = head;return;}if(mp.count(key) > 0) {ListNode1 *curNode = mp[key];curNode->val = value;if(curNode == head)return;curNode->pre->next = curNode->next;curNode->next->pre = curNode->pre;head->pre->next = curNode;curNode->pre = head->pre;curNode->next = head;head->pre = curNode;head = curNode;return;}if(size < capacity) {size ++;ListNode1 *newNode = new ListNode1(key, value);mp[key] = newNode;if(size == 1) {newNode->pre = newNode;newNode->next = newNode;head = newNode;return;}head->pre->next = newNode;newNode->pre = head->pre;newNode->next = head;head->pre = newNode;head = newNode;return;}// size >= capacityListNode1 *tail = head->pre;mp.erase(mp.find(tail->key));mp[key] = tail;tail->key = key;tail->val = value;head = tail;} private:int capacity;int size;map<int, ListNode1*> mp;ListNode1 *head; };

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轉載于:https://www.cnblogs.com/xinsheng/p/3514968.html

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