LeetCode 0146. LRU 缓存：双向链表 + 哈希

【LetMeFly】146.LRU 缓存：双向链表 + 哈希

力扣题目链接：https://leetcode.cn/problems/lru-cache/

函数 get 和 put 必须以 O(1) 的平均时间复杂度运行。

 

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 <= 10000
• 0 <= value <= 105
• 最多调用 2 * 105 次 get 和 put

方法一：双向链表 + 哈希

使用一个双向链表来作为LRU缓存。越靠近链表头部的节点使用时间越近。

使用一个哈希表，来实现从key映射到节点的功能。

为了能从节点映射到哈希表的键值key，在节点中也额外存储一份这个节点的key值：

class LRU_Node {
public:
    LRU_Node* previous, *next;
    int key, value;
}
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为了方便操作，可以在双向链表的首尾各添加一个空节点，以避免“是否为空”的特判。

对于get操作：

若哈希表中存有该key，则由哈希表映射出该节点，将该节点移动为链表的第一个节点，并返回节点的value。

若哈希表中不存在该key，直接返回-1。

对于put操作：

若哈希表中存有该key，则由哈希表映射出该节点，更新该节点的值，并将该节点移动为链表的第一个节点。

若哈希表中不存在该key，创建该节点并将其置于链表的第一个节点。若哈希表的容量大于最大容量，则由tail.previous得到最后一个节点，在哈希表中删除这个节点的key，并在链表中删除这个节点。

• 时间复杂度：每次操作的时间复杂度都是$O(1)$
• 空间复杂度$O(max(put,capacity))$

AC代码

C++

class LRU_Node {
public:
    LRU_Node* previous, *next;
    int key, value;
    LRU_Node(LRU_Node* previous, LRU_Node* next, int key, int value) {
        this->previous = previous;
        this->next = next;
        this->key = key;
        this->value = value;
    }
};

class LRUCache {
private:
    LRU_Node* head, *tail;
    int capacity;
    unordered_map<int, LRU_Node*> ma;

    void refresh(int key, int value) {
        LRU_Node* thisNode = ma[key];
        thisNode->value = value;

        LRU_Node* previous = thisNode->previous, *next = thisNode->next;
        previous->next = next, next->previous = previous;
        
        thisNode->next = head->next;
        head->next = thisNode;
        thisNode->previous = head;
        thisNode->next->previous = thisNode;
    }
public:
    LRUCache(int capacity) {
        head = new LRU_Node(nullptr, nullptr, 0, 0);
        tail = new LRU_Node(head, nullptr, 0, 0);
        head->next = tail;
        this->capacity = capacity;
    }
    
    int get(int key) {
        if (ma.count(key)) {
            refresh(key, ma[key]->value);
            return ma[key]->value;
        }
        return -1;
    }
    
    void put(int key, int value) {
        if (ma.count(key)) {
            refresh(key, value);
            return;
        }
        LRU_Node* thisNode = new LRU_Node(head, head->next, key, value);
        ma[key] = thisNode;
        head->next = thisNode, thisNode->next->previous = thisNode;
        if (ma.size() > capacity) {
            LRU_Node* toRemove = tail->previous;
            ma.erase(toRemove->key);
            toRemove->previous->next = tail;
            tail->previous = toRemove->previous;
        }
    }

    void debug() {
        cout << "Now size: " << ma.size() << ": [";
        LRU_Node* p = head->next;
        while (p != tail) {
            if (p != head->next) {
                cout << ", ";
            }
            cout << "(" << p->key << "|" << p->value << ")";
            p = p->next;
        }
        cout << "] | [";
        p = tail->previous;
        while (p != head) {
            if (p != tail->previous) {
                cout << ", ";
            }
            cout << "(" << p->key << "|" << p->value << ")";
            p = p->previous;
        }
        cout << "]" << endl;
    }
};
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Python

class LRU_Node:
    
    def __init__(self, previous, next, key, value):
        self.previous = previous
        self.next = next
        self.key = key
        self.value = value


class LRUCache:

    def __init__(self, capacity: int):
        self.capacity = capacity
        self.head = LRU_Node(None, None, 0, 0)
        self.tail = LRU_Node(self.head, None, 0, 0)
        self.head.next = self.tail
        self.ma = dict()
    

    def move2first(self, thisNode: LRU_Node):
        thisNode.previous.next = thisNode.next
        thisNode.next.previous = thisNode.previous
        
        thisNode.previous = self.head
        thisNode.next = self.head.next
        self.head.next = thisNode
        thisNode.next.previous = thisNode


    def get(self, key: int) -> int:
        if key in self.ma:
            self.move2first(self.ma[key])
            return self.ma[key].value
        return -1


    def put(self, key: int, value: int) -> None:
        if key in self.ma:
            thisNode = self.ma[key]
            thisNode.value = value
            self.move2first(thisNode)
        else:
            thisNode = LRU_Node(self.head, self.head.next, key, value)
            self.ma[key] = thisNode
            self.head.next = thisNode
            thisNode.next.previous = thisNode
            if len(self.ma) > self.capacity:
                toRemove = self.tail.previous
                del self.ma[toRemove.key]
                toRemove.previous.next = self.tail
                self.tail.previous = toRemove.previous

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