【leetcode】138. Copy List with Random Pointer

题目如下：

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.

Example 1:

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:

1. You must return the copy of the given head as a reference to the cloned list.

解题思路：我的方法是先深拷贝list及其next，同时把新旧两个list的每个元素浅拷贝都存起来放到两个数组中；第二次再遍历旧的list，根据random找到对应item在数组中的下标，再把新的list中同一个item指向数组中对应的下标的元素即可。

代码如下：

"""
# 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
        """
        list1 = []
        list2 = []
        new_head = None
        current = None
        old_head = head
        while old_head is not None:
            if new_head == None:
                new_head = Node(old_head.val,None,None)
                current = new_head
            else:
                new_node = Node(old_head.val,None,None)
                current.next = new_node
                current = current.next
            list1.append(old_head)
            list2.append(current)
            old_head = old_head.next

        old_head = head
        current = new_head
        while old_head is not None:
            if old_head.random is not None:
                inx = list1.index(old_head.random)
                current.random = list2[inx]
            old_head = old_head.next
            current = current.next
        return new_head