64 lines
1.7 KiB
Markdown
64 lines
1.7 KiB
Markdown
---
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id: "202203031"
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date: "2022/03/03 20:38:05"
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title: "leetcode.Q1744.复制带随机指针的链表"
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tags: ["java", "leetcode", "链表"]
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categories:
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- "算法"
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- "leetcode刷题"
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---
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## 解析思路
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leetcode 中等，题目描述[点击这里](https://leetcode-cn.com/problems/copy-list-with-random-pointer/)。
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比较简单，链表结构可以通过遍历复制，关键是random节点如何处理，这里提供一个处理思路：
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<!-- more -->
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通过两次遍历来实现：
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- 第一次遍历建立新的链表，不处理random，遍历的同时将旧链表节点和新链表节点做一个一一对应
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- 第二次遍历处理random节点，根据map找到新链表的位置
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## 代码
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```java
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public class Q138 {
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private static class Node {
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int val;
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Node next;
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Node random;
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public Node(int val) {
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this.val = val;
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this.next = null;
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this.random = null;
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}
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}
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public Node copyRandomList(Node head) {
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//老节点和新节点的对应关系
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Map<Node, Node> oldNewMap = new HashMap<>();
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//建立虚拟头节点便于处理
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Node res = new Node(0), tempNew = res, tempOld = head;
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while (tempOld != null) {
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tempNew.next = new Node(tempOld.val);
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tempNew = tempNew.next;
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//旧节点和新节点一一对应
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oldNewMap.put(tempOld, tempNew);
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tempOld = tempOld.next;
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}
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//再次便利将random关系处理好
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while (head != null) {
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if (head.random != null) {
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oldNewMap.get(head).random = oldNewMap.get(head.random);
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}
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head = head.next;
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}
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return res.next;
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}
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}
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``` |