【leetcode】LRU Cache(hard)★
Design and implement a data structure for Least Recently Used (LRU) cache. It should support the following operations: get
and set
.
get(key)
- Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1.set(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.
思路:
这题吧,思路挺清楚的,就是每次get, set时都把对应的数据设为时间最近的数据,如果满了,就在set时把最老的数据扔掉,把新的插入到最近的位置。
关键是,如何在O(1)的时间内get到所需的数据,如何在O(1)的时间内,找到最老的数据。
第一个问题可以用unordered_map轻松解决,但是,第二个问题我就不会了。我很low的用了遍历,果断超时了。看答案后发现,要用list的splice函数解决。
把所有的数据按照访问时间由近到远存放在一个list中,当再次访问里面的数据时,就把该数据移动到list的开始位置,满了后就移除list的最后一个元素。
上大神的答案:
class LRUCache { private: // A list of (key, value) pairs list<pair<int, int>> items; // Map items to iterators (pointers) to list nodes unordered_map<int, list<pair<int, int>>::iterator> cache; // The capacity of the list int capacity; public: LRUCache(int capacity) : capacity(capacity) {} int get(int key) { // If key is not found in hash map, return -1 if (cache.find(key) == cache.end()) return -1; // Move the (key, value) pair to the beginning of the list items.splice(items.begin(), items, cache[key]); return cache[key]->second; } void set(int key, int value) { // The key is not in the hash table if (cache.find(key) == cache.end()) { // If the cache is full then delete the least recently // used item, which is at the end of the list if (items.size() == capacity) { cache.erase(items.back().first); items.pop_back(); } items.push_front(make_pair(key, value)); cache[key] = items.begin(); } else { // Update the value associated with the key cache[key]->second = value; // Move the (key, value) pair to the beginning of the list items.splice(items.begin(), items, cache[key]); } } }
我的代码,时间是用自己设的time来记录的,超时了。
typedef struct Data { int value; int time; Data(){} Data(int v, int t) : value(v), time(t){} }Data; class LRUCache{ public: LRUCache(int capacity) { t = 0; //初始化时间 c = capacity; //初始化容量 } int get(int key) { unordered_map<int, Data>::iterator it = record.find(key); if(it == record.end()) { return -1; } else { it->second.time = t++; return it->second.value; } } void set(int key, int value) { if(record.find(key) != record.end()) { record[key].value = value; record[key].time = t++; return; } if(record.size() == c) //容量已经达到 { unordered_map<int, Data>::iterator replace = record.begin(); for(unordered_map<int, Data>::iterator it = record.begin(); it != record.end(); it++) { replace = (it->second.time < replace->second.time) ? it : replace; } record.erase(replace); //删掉时间最早的 } Data newData(value, t); record[key] = newData; t++; } private: unordered_map<int, Data> record; int c; int t; };
郑重声明:本站内容如果来自互联网及其他传播媒体,其版权均属原媒体及文章作者所有。转载目的在于传递更多信息及用于网络分享,并不代表本站赞同其观点和对其真实性负责,也不构成任何其他建议。