leetcode 146. LRU Cache
2017.04.25
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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?
class LRUCache {
// 基本思想是用一个 map 保存所有的对,用于查找,
// 用一个双向列表维持顺序。
int m_capacity;
unordered_map<int, list<pair<int, int>>::iterator> m_map; //m_map_iter->first: key, m_map_iter->second: list iterator;
list<pair<int, int>> m_list; //m_list_iter->first: key, m_list_iter->second: value;
public:
LRUCache(int capacity):m_capacity(capacity) {
}
int get(int key) {
auto found_iter = m_map.find(key);
if (found_iter == m_map.end()) //key doesn't exist
return -1;
m_list.splice(m_list.begin(), m_list, found_iter->second); //move the node corresponding to key to front
return found_iter->second->second;
}
void put(int key, int value) {
auto found_iter = m_map.find(key);
if (found_iter != m_map.end()) //key exists
{
m_list.splice(m_list.begin(), m_list, found_iter->second); //move the node corresponding to key to front
found_iter->second->second = value; //update value of the node
return;
}
if (m_map.size() == m_capacity) //reached capacity
{
int key_to_del = m_list.back().first;
m_list.pop_back(); //remove node in list;
m_map.erase(key_to_del); //remove key in map
}
m_list.emplace_front(key, value); //create new node in list
m_map[key] = m_list.begin(); //create correspondence between key and node
}
};
/**
* 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);
*/