To prove safety offer-- complex chain of copy

A complex input list (each node has a node value, and two pointers, one pointing to the next node, a special pointer to any other node),
returns a value after the head of the list replication complex. (Note that the output results do not return parameter node reference, otherwise the program will return empty sentenced questions directly)

Problem-solving ideas: both sides of the cycle:

1: two first pass opening arrays, the presence of each of the first node, the second node exists for each value of the new node is opened. While establishing a node memory address to index mapping id2index

2: to open a new traversing node array, and the next supplement by id2index random pointers.

# class RandomListNode:
#     def __init__(self, x):
#         self.label = x
#         self.next = None
#         self.random = None
class Solution:
    # 返回 RandomListNode
    def Clone(self, pHead):
        # write code here
        if pHead==None:
            return None
        iter_array,array=[],[]
        id2index=dict()
        index=0
        #p=pHead
        while pHead:
            iter_array.append(pHead)
            array.append(RandomListNode(pHead.label))
            id2index[id(pHead)]=index
            index+=1
            pHead=pHead.next
        for p, res in zip(iter_array,array):
            if p.next:
                res.next=array[id2index[id(p.next)]]
            if p.random:
                res.random=array[id2index[id(p.random)]]
        return array[0]