146. LRU Cache

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146. LRU Cache

Description

Difficulty: Medium

Related Topics: Hash Table, Linked List, Design, Doubly-Linked List

Design a data structure that follows the constraints of a Least Recently Used (LRU) cache.

Implement the LRUCache class:

  • LRUCache(int capacity) Initialize the LRU cache with positive size capacity.
  • int get(int key) Return the value of the key if the key exists, otherwise return -1.
  • void put(int key, int value) Update the value of the key if the key exists. Otherwise, add the key-value pair to the cache. If the number of keys exceeds the capacity from this operation, evict the least recently used key.

The functions get and put must each run in O(1) average time complexity.

Example 1:

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Input
["LRUCache", "put", "put", "get", "put", "get", "put", "get", "get", "get"]
[[2], [1, 1], [2, 2], [1], [3, 3], [2], [4, 4], [1], [3], [4]]
Output
[null, null, null, 1, null, -1, null, -1, 3, 4]

Explanation
LRUCache lRUCache = new LRUCache(2);
lRUCache.put(1, 1); // cache is {1=1}
lRUCache.put(2, 2); // cache is {1=1, 2=2}
lRUCache.get(1);    // return 1
lRUCache.put(3, 3); // LRU key was 2, evicts key 2, cache is {1=1, 3=3}
lRUCache.get(2);    // returns -1 (not found)
lRUCache.put(4, 4); // LRU key was 1, evicts key 1, cache is {4=4, 3=3}
lRUCache.get(1);    // return -1 (not found)
lRUCache.get(3);    // return 3
lRUCache.get(4);    // return 4

Constraints:

  • 1 <= capacity <= 3000
  • 0 <= key <= 104
  • 0 <= value <= 105
  • At most 2 * 105 calls will be made to get and put.

Hints/Notes

Solution

Language: C++

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class LRUCache {
public:
    int size;
    list<pair<int, int>> l;
    map<int, list<pair<int, int>>::iterator> m;


    LRUCache(int capacity) {
        size = capacity;
    }

    void add_to_end(int key, int value) {
        l.push_back({key, value});
        m[key] = next(l.end(), -1);
    }

    int get(int key) {
        if (!m.count(key)) {
            return -1;
        } else {
            list<pair<int,int>>::iterator i = m[key];
            int value = (*i).second;
            l.erase(i);
            add_to_end(key, value);
            return value;
        }
    }

    void put(int key, int value) {
        if (!m.count(key)) {
            if (l.size() == size) {
                int k = l.front().first;
                m.erase(k);
                l.pop_front();
            }
            add_to_end(key, value);
        } else {
            auto i = m[key];
            l.erase(i);
            add_to_end(key, value);
        }
    }
};

/**
 * 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);
 */