• Total Accepted: 109086
• Total Submissions: 675208
• Difficulty: Hard
• Contributors: Admin

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

 We need two data structures：

　　list: list<pair<int, int>>//record the order of objects

　　hashmap: map<int, list<pair<int, int>>::iterator>//mapping the relationship between key and objects

　　　　　　　　　　　　　　　　　　　　　　　　　　//must use the iterator, otherwise the splice function will not work

Functions:

Set(key, value)//If the object exists int the map, erase it in the list first, and then push the object from the front. Set m[key] = l.front();

　　//if the size of map is larger than capacity, erase the object both from list and map;

Get(key)//If the objects do not exist in map, return -1;

　　//The only change is the order in the list.

　　//to_list.splice(position, from_list, iterator)

 1 class LRUCache {
 2 public:
 3     LRUCache(int capacity) {
 4         cap = capacity;
 5     }
 6     
 7     int get(int key) {
 8         auto it = m.find(key);
 9         if(it == m.end()) return -1;//not find
10         l.splice(l.begin(), l, it -> second);//move it in l to l.begin();
11         return it -> second -> second;
12     }
13     
14     void put(int key, int value) {
15         auto it = m.find(key);
16         if(it != m.end()) l.erase(it -> second);
17         //l.push_front(it -> second);
18         l.push_front(make_pair(key, value));
19         m[key] = l.begin();
20         if(m.size() > cap){
21             auto k = l.rbegin() -> first;//要存int类型 不然指针unsafe 因为后面list要pop_back
22             l.pop_back();
23             //m.erase(k -> first);
24             m.erase(k);
25         }
26     }
27 private:
28     int cap;
29     list<pair<int, int>> l;//record the order of least recently used
30     map<int, list<pair<int, int>>::iterator> m;//hashmap 一定要iterator
31 };
32 
33 /**
34  * Your LRUCache object will be instantiated and called as such:
35  * LRUCache obj = new LRUCache(capacity);
36  * int param_1 = obj.get(key);
37  * obj.put(key,value);
38  */

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