题目
题目链接:
https://www.nowcoder.com/practice/5dfded165916435d9defb053c63f1e84
思路
双向链表+map
最新的数据放头结点,尾节点放最老的数据,没次移除尾巴节点
本地考察链表的新增,删除,移动节点
参考答案Java
import java.util.*;
public class Solution {
Map<Integer, Node> cache = new HashMap<>();
Node start, end;
int cap = 0;
public Solution(int capacity) {
// write code here
cap = capacity;
}
public int get(int key) {
//key对应节点移动到头部,成为头节点
if (!cache.containsKey(key)) return -1;
Node cur = cache.get(key);
int v = cur.data;
Node next = cur.next;
Node prev = cur.prev;
if (next != null && prev != null) { //cur 要变成头结点
next.prev = prev;
prev.next = next;
if (next.next == null) { //这里似乎可以不要
end = next;
}
cur.next = start;
start.prev = cur;
start = cur;
} else if (next != null) { //说明cur是头结点,不管了
} else if (prev != null) { //自己是尾结点
prev.next = null; //自己的prev要成为尾巴,prev.next设置为null
cur.next = start;
start.prev = cur;
start = cur;
end = prev; //尾巴修改为自己的前一个节点
}
return v;
}
public void set(int key, int value) {
if (cache.containsKey(key)) {
cache.get(key).data = value;
cache.put(key, cache.get(key));
get(key); //使用一次,移动到头部
} else {
Node node = new Node(key, value);
if (cap == 1) { //容量为1时特殊处理
start = end = node;
cache.clear();
cache.put(key, node);
return;
}
int size = cache.size();
if (start == null) {
start = node;
end = node;
cache.put(key, node);
} else if (size < cap) { //不需要移除尾节点,直接修改头部
node.next = start;
start.prev = node;
start = node;
cache.put(key, node);
} else {
// System.out.println();
// System.out.println(key+" == "+value);
// System.out.println();
Node last = end;
Node lastprev = last.prev;
end = lastprev; //设置新的尾节点
cache.remove(last.key);
end.next = null;
last = null;
node.next = start;
start.prev = node;
start = node; //设置新的头结点
cache.put(key, node);
}
//show(start);
}
}
static class Node {
int key;
int data;
Node prev;
Node next;
public Node(int k, int d) {
key = k;
data = d;
}
}
public void show(Node root) { //帮助打印的,本答案可以不需要
System.out.println("");
Node t = root;
Set<Integer> s = new HashSet<>();
while (t != null) {
System.out.print(t.key + "=>" + t.data + " ");
t = t.next;
//if(s.contains(t.data)) break;
}
System.out.println("");
}
}
/**
* Your Solution object will be instantiated and called as such:
* Solution solution = new Solution(capacity);
* int output = solution.get(key);
* solution.set(key,value);
*/
本答案在lintcode 上相同题目没有通过全部测试用例
https://www.lintcode.com/problem/134/
后期找到原因后再修改本答案