LeetCode_146.LRU 缓存机制

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进阶：你是否可以在 O(1) 时间复杂度内完成这两种操作？

示例：

输入
["LRUCache", "put", "put", "get", "put", "get", "put", "get", "get", "get"]
[[2], [1, 1], [2, 2], [1], [3, 3], [2], [4, 4], [1], [3], [4]]
输出
[null, null, null, 1, null, -1, null, -1, 3, 4]

解释
LRUCache lRUCache = new LRUCache(2);
lRUCache.put(1, 1); // 缓存是 {1=1}
lRUCache.put(2, 2); // 缓存是 {1=1, 2=2}
lRUCache.get(1);    // 返回 1
lRUCache.put(3, 3); // 该操作会使得关键字 2 作废，缓存是 {1=1, 3=3}
lRUCache.get(2);    // 返回 -1 (未找到)
lRUCache.put(4, 4); // 该操作会使得关键字 1 作废，缓存是 {4=4, 3=3}
lRUCache.get(1);    // 返回 -1 (未找到)
lRUCache.get(3);    // 返回 3
lRUCache.get(4);    // 返回 4

提示：

• 1 <= capacity <= 3000
• 0 <= key <= 3000
• 0 <= value <= 104
• 最多调用 3 * 104 次 get 和 put

C#代码

public class LRUCache
{
    System.Collections.Concurrent.ConcurrentDictionary<int, LinkedListNode<Tuple<int, int>>> dic;
    LinkedList<Tuple<int, int>> linkedList;
    int cap;

    public LRUCache(int cap)
    {
        dic = new System.Collections.Concurrent.ConcurrentDictionary<int, LinkedListNode<Tuple<int, int>>>();
        this.linkedList = new LinkedList<Tuple<int, int>>();
        this.cap = cap;
    }

    public int Get(int key)
    {
        dic.TryGetValue(key, out LinkedListNode<Tuple<int, int>> val);
        if (val == null) return -1;

        int value = val.Value.Item2;
        Put(key, value);
        return value;
    }

    public void Put(int key, int value)
    {
        dic.TryGetValue(key, out LinkedListNode<Tuple<int, int>> val);
        if (val == null && dic.Count == cap) val = linkedList.Last;

        if (val != null)
        {
            dic.TryRemove(val.Value.Item1,out val);
            linkedList.Remove(val);
        }

        val = new LinkedListNode<Tuple<int, int>>(Tuple.Create(key, value));
        dic.TryAdd(key, val);
        linkedList.AddFirst(val);
    }
}

LeetCode_146.LRU 缓存机制

标签：完成   origin   count   new   http   amp   缓存   ons   限时   

原文地址：https://www.cnblogs.com/fuxuyang/p/14242735.html