13.19 LRU Cache
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2023-12-01
Question
- leetcode: LRU Cache | LeetCode OJ
- lintcode: (134) LRU Cache
Problem Statement
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.
题解
Java
public class Solution {
private int capacity;
private HashMap<Integer, Node> map = new HashMap<>();
private Node head = new Node(-1, -1), tail = new Node(-1, -1);
private class Node {
Node prev, next;
int val, key;
public Node(int key, int val) {
this.val = val;
this.key = key;
prev = null;
next = null;
}
// @Override
// public String toString() {
// return "(" + key + ", " + val + ") " + "last:"
// + (prev == null ? "null" : "node");
// }
}
public Solution(int capacity) {
this.capacity = capacity;
tail.prev = head;
head.next = tail;
}
public int get(int key) {
if (!map.containsKey(key)) {
return -1;
}
// remove current
Node currentNode = map.get(key);
currentNode.prev.next = currentNode.next;
currentNode.next.prev = currentNode.prev;
// move current to tail;
moveToTail(currentNode);
return map.get(key).val;
}
public void set(int key, int value) {
if (get(key) != -1) {
map.get(key).val = value;
return;
}
if (map.size() == capacity) {
map.remove(head.next.key);
head.next = head.next.next;
head.next.prev = head;
}
Node insert = new Node(key, value);
map.put(key, insert);
moveToTail(insert);
}
private void moveToTail(Node current) {
current.prev = tail.prev;
tail.prev = current;
current.prev.next = current;
current.next = tail;
}
}