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Solution_146.java
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121 lines (106 loc) · 3.65 KB
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/*
146. LRU缓存机制
运用你所掌握的数据结构,设计和实现一个 LRU (最近最少使用) 缓存机制。它应该支持以下操作: 获取数据 get 和 写入数据 put 。
获取数据 get(key) - 如果密钥 (key) 存在于缓存中,则获取密钥的值(总是正数),否则返回 -1。
写入数据 put(key, value) - 如果密钥已经存在,则变更其数据值;如果密钥不存在,则插入该组「密钥/数据值」。当缓存容量达到上限时,它应该在写入新数据之前删除最久未使用的数据值,从而为新的数据值留出空间。
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:
LRUCache cache = new LRUCache( 2 ); //缓存容量
cache.put(1, 1);
cache.put(2, 2);
cache.get(1); // 返回 1
cache.put(3, 3); // 该操作会使得密钥 2 作废
cache.get(2); // 返回 -1 (未找到)
cache.put(4, 4); // 该操作会使得密钥 1 作废
cache.get(1); // 返回 -1 (未找到)
cache.get(3); // 返回 3
cache.get(4); // 返回 4
*/
/*
["LRUCache","get","put","get","put","put","get","get"]
[[2],[2],[2,6],[1],[1,5],[1,2],[1],[2]]
*/
import java.util.*;
public class Solution {
public static void main(String[] args) {
LRUCache cache = new LRUCache(2); //缓存容量
System.out.print("cache.get(2): ");
System.out.println(cache.get(2));
System.out.println("cache.put(2,6);");
cache.put(2, 6);
System.out.print("cache.get(1): ");
System.out.println(cache.get(1));
System.out.println("cache.put(1,5);");
cache.put(1, 5);
System.out.println("cache.put(1,2);");
cache.put(1, 2);
System.out.print("cache.get(1): ");
System.out.println(cache.get(1));
System.out.print("cache.get(2): ");
System.out.println(cache.get(2));
}
}
class DeLinkedList {
int val;
int key;
DeLinkedList next;
DeLinkedList pre;
public DeLinkedList(int x, int y) {
key = x;
val = y;
}
}
class LRUCache {
int count = 0;
int max;
DeLinkedList head = new DeLinkedList(-1, -1);
DeLinkedList tail = new DeLinkedList(-1, -1);
Map<Integer, DeLinkedList> hash = new HashMap<>();
public LRUCache(int capacity) {
max = capacity;
head.next = tail;
tail.pre = head;
}
public int get(int key) {
DeLinkedList cur = hash.getOrDefault(key, head);
if (cur.val != -1) {
cur.pre.next = cur.next;
cur.next.pre = cur.pre;
cur.next = head.next;
head.next.pre = cur;
head.next = cur;
cur.pre = head;
}
return cur.val;
}
public void put(int key, int value) {
if (hash.containsKey(key)) {
DeLinkedList p = hash.get(key);
p.val = value;
if (p.pre != head) {
p.pre.next = p.next;
p.next.pre = p.pre;
p.next = head.next;
head.next.pre = p;
head.next = p;
p.pre = head;
}
} else {
DeLinkedList n = new DeLinkedList(key, value);
hash.put(key, n);
if (count >= max) {
DeLinkedList t = tail.pre;
tail.pre.pre.next = tail;
tail.pre = tail.pre.pre;
hash.remove(t.key, t);
count--;
}
n.next = head.next;
head.next.pre = n;
head.next = n;
n.pre = head;
count++;
}
}
}