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LeetCode OJ - LRU Cache

2024-07-04 08:51:41   224人阅读

题目：　

　　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.

解题思路：

　　每一个Cache行有三个域（key, value, update_cnt)；

　　用unordered_map来存储Cache体；

　　用queue来保存使用的历史信息。

代码：

　　

struct Line {
    int key;
    int value;
    int update_cnt;
};

class LRUCache{
public:
    int capacity;
    unordered_map<int, Line> cache;
    queue<Line> q;
    LRUCache(int capacity) {
        this->capacity = capacity;
    }

    int get(int key) {
        if (cache.count(key)) {
            cache[key].update_cnt++;
            q.push(cache[key]);
            return cache[key].value;
        }
        return -1;
    }

    void set(int key, int value) {
        if (cache.count(key)) {
            cache[key].value = value;
            cache[key].update_cnt++;
            q.push(cache[key]);
        }
        else {
            if (cache.size() == capacity) {
                Line tmp;
                while (!q.empty()) {
                    tmp = q.front();
                    q.pop();

                    if (cache.count(tmp.key) && tmp.update_cnt == cache[tmp.key].update_cnt) {
                        break;
                    }
                }
                cache.erase(cache.find(tmp.key));
                cache[key].key = key;
                cache[key].value = value;
                cache[key].update_cnt = 0;
                q.push(cache[key]);
                return;
            }
            cache[key].key = key;
            cache[key].value = value;
            cache[key].update_cnt = 0;
            q.push(cache[key]);
        }
    }
};

 


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