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No146_LRUCache.cpp
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// 146. LRU Cache
// 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
//难点在于如果在O(1)时间内实现Put操作。如果是单一的数据结构,就我的知识来说是无法完成这个要求的
//因此想到空间换时间,结合俩个数据结构来实现
class LRUCache {
public:
LRUCache(int capacity) {
cap = capacity;
}
int get(int key) {
if(!hash_map.count(key)) return -1;
// If find key, modify the order of cache
auto it = hash_map[key];
cache.splice(cache.begin(), cache, it); // Put it into front, hash_map[key] dont't change, still point to the origin
return it->second;
}
void put(int key, int value) {
// If the key already in hash_map
if(hash_map.count(key)){
cache.erase(hash_map[key]);
}
cache.push_front(make_pair(key, value));
hash_map[key] = cache.begin();
// Capacity
if(cache.size() > cap){
auto last = cache.back().first;
cache.pop_back();
hash_map.erase(last);
}
}
int cap; // Capacity of cache
list<pair<int, int> > cache; // Cache
unordered_map<int, list<pair<int,int> >::iterator> hash_map; // Store key and its location in cache
};
/**
* 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);
*/