Leetcode之算法专题《LRU Cache》

题目内容如下（链接：https://leetcode.com/problems/lru-cache/description/）

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

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.

Follow up:
Could you do both operations in O(1) time complexity?

Example:

LRUCache cache = new LRUCache( 2 /* capacity */ );

cache.put(1, 1);
cache.put(2, 2);
cache.get(1);       // returns 1
cache.put(3, 3);    // evicts key 2
cache.get(2);       // returns -1 (not found)
cache.put(4, 4);    // evicts key 1
cache.get(1);       // returns -1 (not found)
cache.get(3);       // returns 3
cache.get(4);       // returns 4

经典面试算法题，用list+unordered_map即可轻松实现，代码如下：

class LRUCache {
private:
    typedef list::iterator IT;
    int m_iCapacity;
    list m_list;
    unordered_map > m_umap;
public:
    LRUCache(int capacity) {
        m_iCapacity = capacity;
        m_list = list();
        m_umap = unordered_map >(); 
    }
    
    int get(int key) {
        if (m_umap.find(key) == m_umap.end())
        {
            return -1;
        }
        
        m_list.erase(m_umap[key].second);
        m_list.push_front(key);
        m_umap[key].second = m_list.begin();
        return m_umap[key].first;
    }
    
    void put(int key, int value) {
        // the key already exist
        if(m_umap.find(key) != m_umap.end())
        {
            m_list.erase(m_umap[key].second);
            m_umap.erase(key);
        }
        else if(m_iCapacity == m_list.size())
        {
            m_umap.erase(m_list.back());
            m_list.pop_back();
        }
        m_list.push_front(key);
        m_umap[key] = pair(value, m_list.begin());
    }
};