复杂链表的复制 牛客网 剑指Offer

复杂链表的复制 牛客网 剑指Offer

• 题目描述
• 输入一个复杂链表（每个节点中有节点值，以及两个指针，一个指向下一个节点，另一个特殊指针指向任意一个节点），返回结果为复制后复杂链表的head。（注意，输出结果中请不要返回参数中的节点引用，否则判题程序会直接返回空）

# class RandomListNode:
#     def __init__(self, x):
#         self.label = x
#         self.next = None
#         self.random = None
class Solution:
    #run:32ms memory:5728k
    def Clone(self, pHead):
        if pHead == None:
            return 
        head = pHead
        while head:
            nd = RandomListNode(0)
            nd.label = head.label
            nd.next = head.next
            head.next = nd
            head = nd.next
            
        head = pHead
        while head:
            nd = head.next
            if head.random != None:
                nd.random = head.random.next
            head = nd.next
        
        head = pHead
        nHead =  head.next
        nNode = head.next
        head.next = nNode.next
        head = head.next
        while head :
            nNode.next = head.next
            nNode = nNode.next
            head.next = nNode.next
            head = head.next
        return nHead