请你设计并实现一个满足 LRU (最近最少使用) 缓存 约束的数据结构。
实现 LRUCache 类:
LRUCache(int capacity)以 正整数 作为容量capacity初始化 LRU 缓存int get(int key)如果关键字key存在于缓存中,则返回关键字的值,否则返回-1。void put(int key, int value)如果关键字key已经存在,则变更其数据值value;如果不存在,则向缓存中插入该组key-value。如果插入操作导致关键字数量超过capacity,则应该 逐出 最久未使用的关键字。
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 <= 30000 <= key <= 100000 <= value <= 105- 最多调用
2 * 105次get和put
解题答案
class LRUCache {
private Map<Integer, Node> map;
private final int capacity;
private final Node first;
private final Node last;
public LRUCache(int capacity) {
map = new HashMap<>(capacity);
this.capacity = capacity;
first = new Node();
last = new Node();
first.next = last;
last.prev = first;
}
public int get(int key) {
Node node = map.get(key);
if (node == null) return -1;
removeNode(node);
addAfterFirst(node);
return node.value;
}
public void put(int key, int value) {
Node node = this.map.get(key);
if (node != null) {
node.value = value;
removeNode(node);
} else {
if (map.size() >= this.capacity) {
removeNode(map.remove(last.prev.key));
}
node = new Node(key, value);
map.put(key, node);
}
addAfterFirst(node);
}
private void removeNode(Node node) {
node.prev.next = node.next;
node.next.prev = node.prev;
}
private void addAfterFirst(Node node) {
node.next = first.next;
node.next.prev = node;
node.prev = first;
first.next = node;
}
static class Node {
int key;
int value;
Node prev;
Node next;
Node(int key, int value) {
this.key = key;
this.value = value;
}
Node() {
}
}
}
/**
* Your LRUCache object will be instantiated and called as such:
* LRUCache obj = new LRUCache(capacity);
* int param_1 = obj.get(key);
* obj.put(key,value);
*/