LRU Cache

Desicription

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:

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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

Solution

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class LRUCache {
private:
int _capacity;
list<int> used;
unordered_map<int, pair<int, list<int>::iterator>> cache;
void touch(unordered_map<int, pair<int, list<int>::iterator>>::iterator it) {
used.erase(it->second.second);
used.push_front(it->first);
it->second.second = used.begin();
}
public:
LRUCache(int capacity) : _capacity(capacity) {};
int get(int key) {
auto it = cache.find(key);
if(it == cache.end())
return -1;
touch(it);
return it->second.first;
}
void put(int key, int value) {
auto it = cache.find(key);
if(it != cache.end())
touch(it);
else {
if(cache.size() == _capacity) {
cache.erase(used.back());
used.pop_back();
}
used.push_front(key);
}
cache[key] = {value, used.begin()};
}
};