标签：iterator lex 迭代 else and cap complex sts ==
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.
The cache is initialized with a positive capacity.
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
class LRUCache { unordered_map<uint32_t,pair<list<uint32_t>::iterator,uint32_t>> cache; list<uint32_t> order; int capacity; public: LRUCache(int capacity) { this->capacity = capacity; } int get(int key) { auto iter=cache.find(key); if(iter==cache.end())return -1; update(iter); return iter->second.second; } void update(unordered_map<uint32_t,pair<list<uint32_t>::iterator,uint32_t>>::iterator i) { order.erase(i->second.first); order.push_front(i->first); i->second.first=order.begin(); } void put(int key, int value) { auto iter=cache.find(key); if(iter==cache.end()) { if(cache.size()>=capacity) { cache.erase(order.back()); order.pop_back(); } order.push_front(key); cache.insert(make_pair(key,make_pair(order.begin(),value))); } else { iter->second.second=value; update(iter); } } };
标签：iterator lex 迭代 else and cap complex sts ==
原文地址：https://www.cnblogs.com/lychnis/p/11873887.html
