leetcode--138. Copy List with Random Pointer

Posted 2020-10-11 Shihu

1，问题描述

A linked list is given such that each node contains an additional random pointer which could point to any node in the list or null.

Return a deep copy of the list.

数据结构：

1 /**
2  * Definition for singly-linked list with a random pointer.
3  * class RandomListNode {
4  *     int label;
5  *     RandomListNode next, random;
6  *     RandomListNode(int x) { this.label = x; }
7  * };
8  */

2，边界条件：root==null情况可以处理，所以没有边界条件，不需要特殊处理

3，解题思路：先不管random指针，把链表节点依次复制，不过不是生成一个新的链表，而是把新节点放在原节点的next。然后再根据random指针把新节点的random指针指向原节点的next，即新节点。最后把大链表分离，恢复旧链表，把新节点生成一个新链表，即为旧链表的deep copy。

4，代码实现

 1 public RandomListNode copyRandomList(RandomListNode head) {
 2     if (head == null) {
 3         return null;
 4     }
 5     RandomListNode cur = head;
 6     while (cur != null) {
 7         RandomListNode node = new RandomListNode(cur.label);
 8         node.next = cur.next;
 9         cur.next = node;
10         cur = node.next;
11     }
12 
13     cur = head;
14     while (cur != null && cur.next != null) {
15         if (cur.random != null) {
16             cur.next.random = cur.random.next;
17         }
18         cur = cur.next.next;
19     }
20 
21     RandomListNode dummy = new RandomListNode(-1);
22     // RandomListNode copyCur = head.next;//这种写法在leetcode不通过，在lintcode上面能通过
23     // dummy.next = copyCur;
24     // cur = head.next.next;
25     RandomListNode copyCur = dummy;//这里是注意下，比上面写法简洁，可以处理head=null的情况
26     cur = head;
27     while (cur != null && cur.next != null) {
28         copyCur.next = cur.next;
29         cur.next = cur.next.next;
30         cur = cur.next;
31         copyCur = copyCur.next;
32     }
33     return dummy.next;
34 }

5，时间复杂度：O（n），空间复杂度：O（1）

6，api：无