LeetCode.146.LRU Cache 实现LRU缓存
题目描述
146 LRU Cache
https://leetcode-cn.com/problems/lru-cache/
Design and implement a data structure for Least Recently Used (LRU) cache. It should support the following operations: get and put.
get(key) - Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1.
put(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.
The cache is initialized with a positive capacity.
Follow up:
Could you do both operations in O(1) time complexity?
Example:
LRUCache cache = new LRUCache( 2 /* capacity */ );
cache.put(1, 1);
cache.put(2, 2);
cache.get(1); // returns 1
cache.put(3, 3); // evicts key 2
cache.get(2); // returns -1 (not found)
cache.put(4, 4); // evicts key 1
cache.get(1); // returns -1 (not found)
cache.get(3); // returns 3
cache.get(4); // returns 4
解题过程
利用Java的LinkedHashMap有序哈希
Java 中,利用 LinkedHashMap
可实现 LRU 缓存,重写其 boolean removeEldestEntry(Map.Entry)
,当容量超过初始容量时返回true即可。
因为 LinkedHashMap 的 putVal()
最后一步 afterNodeInsertion
中当 removeEldestEntry()
返回 true 时会删除 first 结点,如果构造方法中设置了 accessOrder=true
,first结点就是最早访问的结点。
get, put 时间复杂度 O(1)
private static class LRUCache extends LinkedHashMap<Integer, Integer> {
private int capacity;
public LRUCache(int capacity) {
super(capacity, 0.75f, true);
this.capacity = capacity;
}
public int get(int key) {
return super.get(key) != null ? super.get(key) : -1;
}
public void put(int key, int value) {
super.put(key, value);
}
// 重写 removeEldestEntry 当元素个数大于初始容量 capacity 时删除最长时间没访问的Entry
@Override
protected boolean removeEldestEntry(Map.Entry<Integer, Integer> eldest) {
return size() > capacity;
}
}
哈希表+双向链表手动实现
GitHub代码
algorithms/leetcode/leetcode/_146_LRUCache.java
https://github.com/masikkk/algorithms/blob/master/leetcode/leetcode/_146_LRUCache.java
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