leetcode：146. LRU緩存機制

146. LRU緩存機制

思路

一開始的思路是用map來保存每個key和pair<value,順序>，順序是用1，2，3…來表示的，越後面越表示沒被訪問。每當執行get和put的時候，訪問的那個pair就移到1，它之前的頁面就全往後移一位。這樣也是能過簡單的樣例的，但是在最後幾個樣例會超時。因爲它每次put和get都會把所有的map遍歷一遍，非常耗時。

所以就去看了題解，發現需要用到哈希表來解決搜索的問題，用雙向列表來解決排列的問題。這方面詳細的介紹就去看題解吧~

ac代碼

class LRUCache {
public:
    int capacity;
    list<pair<int, int>> l;
    unordered_map<int, list<pair<int, int>>::iterator> hash;

    LRUCache(int capacity) {
        this->capacity = capacity;
    }
    
    int get(int key) {
        auto a = hash.find(key);
        if(a == hash.end()) return -1;
        l.push_front(make_pair(hash[key]->first, hash[key]->second));
        l.erase(a->second);
        hash[key] = l.begin();
        return hash[key]->second;
    }
    
    void put(int key, int value) {
        auto a = hash.find(key);
        if(a != hash.end()){
            l.push_front(make_pair(key, value));
            l.erase(a->second);
            hash[key] = l.begin();
            return;
        }
        if(hash.size() == capacity){
            hash.erase(l.rbegin()->first);
            l.pop_back();
        }
        l.push_front(make_pair(key, value));
        hash[key] = l.begin();
    }
};

/**
 * Your LRUCache object will be instantiated and called as such:
 * LRUCache* obj = new LRUCache(capacity);
 * int param_1 = obj->get(key);
 * obj->put(key,value);
 */

超時代碼

class LRUCache {
public:
    int capacity;
    map<int, pair<int, int>> m;

    LRUCache(int capacity) {
        this->capacity = capacity;
    }
    
    int get(int key) {
        if(m.find(key) == m.end()) return -1;
        for(auto& p: m){
            if(p.second.second < m[key].second)
                p.second.second ++;
        }
        m[key].second = 1;
        return m[key].first;
    }
    
    void put(int key, int value) {
        if(m.find(key) != m.end()){
            for(auto& p: m){
                if(p.second.second < m[key].second)
                    p.second.second ++;
            }
            m[key] = make_pair(value, 1);
            return;
        }
        if(m.size() == capacity){
            for(auto& p: m){
                if(p.second.second == capacity){
                    m.erase(p.first);
                    break;
                }
            }
            for(auto& p: m)
                p.second.second ++;
            m[key] = make_pair(value, 1);
        }
        else{
            for(auto& p: m)
                p.second.second ++;
            m[key] = make_pair(value, 1);
        }
    }
};

/**
 * Your LRUCache object will be instantiated and called as such:
 * LRUCache* obj = new LRUCache(capacity);
 * int param_1 = obj->get(key);
 * obj->put(key,value);
 */