leetcode-460 - LFU Cache

動機

兩題很像，所以就一起寫

Problem

Design and implement a data structure for a Least Frequently Used (LFU) cache.

Implement the LFUCache class:

• LFUCache(int capacity) Initializes the object with the capacity of the data structure.
• int get(int key) Gets the value of the key if the key exists in the cache. Otherwise, returns -1.
• void put(int key, int value) Update the value of the key if present, or inserts the key if not already present. When the cache reaches its capacity, it should invalidate and remove the least frequently used key before inserting a new item. For this problem, when there is a tie (i.e., two or more keys with the same frequency), the least recently used key would be invalidated.

To determine the least frequently used key, a use counter is maintained for each key in the cache. The key with the smallest use counter is the least frequently used key.

When a key is first inserted into the cache, its use counter is set to 1 (due to the put operation). The use counter for a key in the cache is incremented either a get or put operation is called on it.

The functions get and put must each run in O(1) average time complexity.

Example 1:

Input[LFUCache, put, put, get, put, get, get, put, get, get, get][[2], [1, 1], [2, 2], [1], [3, 3], [2], [3], [4, 4], [1], [3], [4]]Output[null, null, null, 1, null, -1, 3, null, -1, 3, 4]Explanation// cnt(x) = the use counter for key x// cache=[] will show the last used order for tiebreakers (leftmost element is  most recent)LFUCache lfu = new LFUCache(2);lfu.put(1, 1);   // cache=[1,_], cnt(1)=1lfu.put(2, 2);   // cache=[2,1], cnt(2)=1, cnt(1)=1lfu.get(1);      // return 1                 // cache=[1,2], cnt(2)=1, cnt(1)=2lfu.put(3, 3);   // 2 is the LFU key because cnt(2)=1 is the smallest, invalidate 2.                 // cache=[3,1], cnt(3)=1, cnt(1)=2lfu.get(2);      // return -1 (not found)lfu.get(3);      // return 3                 // cache=[3,1], cnt(3)=2, cnt(1)=2lfu.put(4, 4);   // Both 1 and 3 have the same cnt, but 1 is LRU, invalidate 1.                 // cache=[4,3], cnt(4)=1, cnt(3)=2lfu.get(1);      // return -1 (not found)lfu.get(3);      // return 3                 // cache=[3,4], cnt(4)=1, cnt(3)=3lfu.get(4);      // return 4                 // cache=[3,4], cnt(4)=2, cnt(3)=3

Constraints:

• 0 <= capacity <= 104
• 0 <= key <= 105
• 0 <= value <= 109
• At most 2 * 105 calls will be made to get and put.

460 LFU Cache

與146不同的是要想辦法記錄每個key的頻率，一定要嘛? 因為當頻率一樣時要變成LRU。

這樣不能像146把所有key放在list中，所以用map紀錄次數，把key放到對應的list中。

為了驅逐的速度上去，最後就是記下如果滿了的話，要驅逐哪一個次數的list的最後一個key

class LFUCache {
private:
    unordered_map<int,list<int>> l;
    int c;
    unordered_map<int, tuple<list<int>::iterator, int, int>> m; // iter, value, times
    int last_frq = 1;
public:
    // new empty list
    tuple<list<int>::iterator, int, int>& moveToTop(int key) {
        auto& tmp = m[key];
        l[std::get<2>(tmp)].erase(std::get<0>(tmp));
        l[std::get<2>(tmp)+1].push_front(key);
        if (l[std::get<2>(tmp)].empty() && last_frq == std::get<2>(tmp))
            last_frq = std::get<2>(tmp)+1;
        auto ret = make_tuple(l[std::get<2>(tmp)+1].begin(), std::get<1>(tmp),std::get<2>(tmp)+1);
        m[key] = ret;
        return m[key];
    }
    LFUCache(int capacity): c(capacity) {
        
    }
    
    int get(int key) {
        auto tmp = m.find(key);
        if(tmp != m.end()) {
            auto& tmp = moveToTop(key);
            return std::get<1>(tmp);
        } else {
            return -1;
        }
    }
    
    void put(int key, int value) {
        auto exist = m.find(key);
        if (exist != m.end()) {
            auto& tmp = moveToTop(key);
            m[key] = make_tuple(std::get<0>(tmp), value, std::get<2>(tmp));
        } else {
            if(m.size() >= c && m.size() > 0) {
                int evict = l[last_frq].back();
                m.erase(evict);
                l[last_frq].pop_back();
                if (l[last_frq].empty()) {
                    l.erase(last_frq);
                    last_frq = l.begin()->first;
                }
                //cout<< "evict: " << evict <<'\n';
            }
            if (m.size() < c) {
                l[1];
                l[1].push_front(key);
                m[key] = make_tuple(l[1].begin(),value, 1);
                last_frq = 1;
            }
        }
        
    }
};