题目描述：
中文：
给定一个链表，每个节点包含一个额外增加的随机指针，该指针可以指向链表中的任何节点或空节点。
要求返回这个链表的深拷贝。
示例：
输入：
{"$id":"1","next":{"$id":"2","next":null,"random":{"$ref":"2"},"val":2},"random":{"$ref":"2"},"val":1}
解释：
节点 1 的值是 1，它的下一个指针和随机指针都指向节点 2 。
节点 2 的值是 2，它的下一个指针指向 null，随机指针指向它自己。
提示：
你必须返回给定头的拷贝作为对克隆列表的引用。
英文：
Return a deep copy of the list.
Input:
{"$id":"1","next":{"$id":"2","next":null,"random":{"$ref":"2"},"val":2},"random":{"$ref":"2"},"val":1}
Explanation:
Node 1's value is 1, both of its next and random pointer points to Node 2.
Node 2's value is 2, its next pointer points to null and its random pointer points to itself.
Note:
You must return the copy of the given head as a reference to the cloned list.
""" # Definition for a Node. class Node(object): def __init__(self, val, next, random): self.val = val self.next = next self.random = random """ class Solution(object): def copyRandomList(self, head): """ :type head: Node :rtype: Node """ nodeDict = dict() dummy = Node(0, None, None) nodeDict[head] = dummy newHead, pointer = dummy, head while pointer: node = Node(pointer.val, pointer.next, None) nodeDict[pointer] = node newHead.next = node newHead, pointer = newHead.next, pointer.next pointer = head while pointer: if pointer.random: nodeDict[pointer].random = nodeDict[pointer.random] pointer = pointer.next return dummy.next
题目来源：力扣