146. LRU Cache

https://leetcode.com/problems/lru-cache

Problem

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.

Follow up: Could you do get and put in O(1) time complexity?

Example 1:

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 <= 3000

• 0 <= value <= 104

• At most 3 * 104 calls will be made to get and put.

Solution

class LRUCache {
private:
    int capacity;
    list<int> keys;
    unordered_map<int, list<int>::iterator> key2idx;
    unordered_map<int, int> key2val;

    inline void updateIdx(int key) {
        keys.erase(key2idx[key]);
        keys.push_front(key);
        key2idx[key] = keys.begin();
    }
    
    inline void evict() {
        const int key2delete = keys.back();
        keys.pop_back();
        key2idx.erase(key2delete);
        key2val.erase(key2delete);
    }
public:
    LRUCache(int cap): capacity(cap) {
        key2idx.reserve(cap);
        key2val.reserve(cap);
    }
    
    int get(int key) {
        if (key2val.find(key) != key2val.end()) {
            updateIdx(key);
            return key2val[key];
        } 
        return -1;
    }
    
    void put(int key, int value) {
        if (key2val.find(key) != key2val.end()) {
            // key exists
            key2val[key] = value;
            updateIdx(key);
        } else {
            // new key
            if (capacity == key2val.size()) evict();
            keys.push_front(key);
            key2idx.insert({key, keys.begin()});
            key2val.insert({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);
 */

• #hash

• #linkedlist