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.

基本思路：

用一个map处理查找问题。

用list维护按使用元素排序。将近期被訪问的记录，总是移动到链表头。

list中存储的是[key, value]，

而map中存储的是key, 以及在list中相应节点的iterator。  

在将list中的元素移动到最前时，使用了splice函数。 此函数比先删除，再插入，要高效。

此代码在leetcode上实际运行时间为160ms。

class LRUCache{public:    LRUCache(int capacity) :capacity_(capacity) {            }        int get(int key) {        auto iter = cache_.find(key);        if (iter == cache_.end())            return -1;                lru_.splice(lru_.begin(), lru_, iter->second);        return iter->second->second;    }        void set(int key, int value) {        if (get(key) != -1) {            lru_.front().second = value;            return;        }                if (lru_.size() == capacity_) {            cache_.erase(lru_.back().first);            lru_.pop_back();        }                lru_.push_front(make_pair(key, value));        cache_[key] = lru_.begin();    }private:    typedef list
     
      
        > list_t;    typedef unordered_map
       
         map_t;    list_t lru_;    map_t cache_;    const int capacity_;};
       
      
     

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