LeetCode——LRU Cache

LRU Cache

 

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.

思路：
最常用的数据结构实现方式是map和Double-Linked List，map只是为了实现键值key-value对应，这样就避免了每次寻找特定值都要在双线链表里面顺序查找，服务器是没办法负担这种低效率的查找方法的。

set流程讲解：

1. 在map中查找key，如果找到，说明这个节点已经存在，那么将该节点的的位置排在链表头部；
2. 如果没找到，要进行第二个判断“是否已经达到了max_size”，如果已经达到了，将原来链表最后一个节点删除，然后将对应的map中的key-value删除，然后再加入新节点，并在map中加入对应的key-value，如果没有达到max-size，那么分配新地址给这个节点，同时加入到list和对应的map中。

get比较简单

代码：

class LRUCache {public:LRUCache(int capacity) {_cache_map.clear();_cache_list.clear();_max_size = capacity;}int get(int key) {map<int, list<pair<int, int> >::iterator>::iterator p = _cache_map.find(key);if (p != _cache_map.end()) {_cache_list.splice(_cache_list.begin(), _cache_list, p->second);p->second = _cache_list.begin();return (p->second)->second;}return -1;}void set(int key, int value) {map<int, list<pair<int, int> >::iterator>::iterator p = _cache_map.find(key);//已经存在keyif (p != _cache_map.end()) {((p->second))->second = value;_cache_list.splice(_cache_list.begin(), _cache_list, p->second);p->second = _cache_list.begin();} else {//key不存在if (_cache_map.size() >= _max_size) {//判断是否达到maxsize_cache_map.erase(_cache_list.rbegin()->first);_cache_list.pop_back();}_cache_list.push_front(pair<int, int>(key, value));_cache_map[key] = _cache_list.begin();}}private:map<int, list<pair<int, int> >::iterator> _cache_map;list<pair<int, int> > _cache_list;int _max_size;};

附：splice是移动链表元素的方法
splice实例代码：

#include<iostream>#include<list>using namespace std;int main() {list<int> mylist1, mylist2;list<int>::iterator it;// set some initial values:for (int i = 1; i <= 4; i++)mylist1.push_back(i);      // mylist1: 1 2 3 4for (int i = 1; i <= 3; i++)mylist2.push_back(i * 10);   // mylist2: 10 20 30it = mylist1.begin();++it;                         // points to 2mylist1.splice(it, mylist2); // mylist1: 1 10 20 30 2 3 4 // mylist2 (empty) // "it" still points to 2 (the 5th element)mylist2.splice(mylist2.begin(), mylist1, it);// mylist1: 1 10 20 30 3 4// mylist2: 2// "it" is now invalid.it = mylist1.begin();advance(it, 3);                // "it" points now to 30mylist1.splice(mylist1.begin(), mylist1, it, mylist1.end());// mylist1: 30 3 4 1 10 20cout << "mylist1 contains:";for (it = mylist1.begin(); it != mylist1.end(); it++)cout << " " << *it;cout << "\nmylist2 contains:";for (it = mylist2.begin(); it != mylist2.end(); it++)cout << " " << *it;cout << endl;return 0;}