146. LRU Cache

146. LRU Cache

题目链接：https://leetcode.com/problems/lru-cache/#/description

题目大意：为最近最少访问cache设计和实现一个数据结构，该数据结构应该支持get和put操作。

　　　　　get(key) - Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1.
　　　　　put(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：因为get和set都涉及频繁的插入删除操作，所以需要用一个list来记录key。同时为了在常数时间访问指定值，因此需要用unordered_map来保存每个key对应的value，同时为了更快地找到被删除的节点，unordered_map也需要保存每个key对应的迭代器。

代码：

class LRUCache {
public:
    LRUCache(int capacity):_capacity(capacity) {
        
    }
    
    int get(int key) {
        auto it = cache.find(key);
        if (it == cache.end())
            return -1;
        touch(it);
        return it->second.first;
    }
    
    void put(int key, int value) {
        auto it = cache.find(key);
        if (it != cache.end())
            touch(it);
        else {
            if (cache.size() == _capacity) {
                cache.erase(keys.back());
                keys.pop_back();
            }
            keys.push_front(key);
        }
        cache[key] = {value, keys.begin()};
    }
private:
    using PII = pair<int, list<int>::iterator>;
    
    void touch(unordered_map<int, PII>::iterator it) {
        keys.erase(it->second.second);
        keys.push_front(it->first);
        it->second.second = keys.begin();
    }
    
    list<int> keys;
    unordered_map<int, PII> cache;
    int _capacity;
};

/**
 * 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);
 */

评测系统上运行结果：