Implement an LRU (Least Recently Used) cache. It should be able to be initialized with cache size, and contain the following methods:
set(key, value)
: setskey
tovalue
. If there are already n items in the cache and we are adding a new item, then it should also remove the least recently used item.get(key)
: gets the value atkey
. If no such key exists, return null.
Each operation should run in O(1) time.
Example:
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
Approach:
C++
#include <bits/stdc++.h>using namespace std;class LRUCache{public:int cap;vector<int> v;unordered_map<int, int> mp;LRUCache(int capacity){cap = capacity;}int get(int key){auto pos = find(v.begin(), v.end(), key);if (pos != v.end()){v.erase(pos);v.push_back(key);return mp[key];}return -1;}void put(int key, int value){auto pos = find(v.begin(), v.end(), key);if (pos != v.end())v.erase(pos);if (v.size() == cap){mp[v[0]] = 0;v.erase(v.begin());}mp[key] = value;v.push_back(key);}};int main(){LRUCache lRUCache(2);lRUCache.put(1, 1);lRUCache.put(2, 2);cout << "[";cout << lRUCache.get(1) << ", ";lRUCache.put(3, 3);cout << lRUCache.get(2) << ", ";lRUCache.put(4, 4);cout << lRUCache.get(1) << ", ";cout << lRUCache.get(3) << ", ";cout << lRUCache.get(4);cout << "]";return 0;}
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