https://www.nowcoder.com/practice/f836b2c43afc4b35ad6adc41ec941dba?tpId=13&tqId=11178&tPage=2&rp=2&ru=%2Fta%2Fcoding-interviews&qru=%2Fta%2Fcoding-interviews%2Fquestion-ranking

題目描述
輸入一個複雜連結串列（每個節點中有節點值，以及兩個指標，一個指向下一個節點，另一個特殊指標指向任意一個節點），返回結果為複製後複雜連結串列的head。（注意，輸出結果中請不要返回引數中的節點引用，否則判題程式會直接返回空）

首先想到的肯定是遞迴來求解了：

# -*- coding:utf-8 -*-

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
        res = RandomListNode(pHead.label)
        res.random = pHead.random
        res.next = self.Clone(pHead.next)
        return res

還一種思路是複製+複製+拆分：

# -*- coding:utf-8 -*-

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

        head = pHead
        while head:
            tmp = head.next
            head.next = RandomListNode(head.label)
            head.next.next = tmp
            head = tmp

        head = pHead
        while head:
            copy_node = head.next
            next_head = copy_node.next
            if head.random:
                copy_node.random = head.random.next
            head = next_head

        head = pHead
        res = pHead.next
        while head:
            copy_node = head.next
            next_head = copy_node.next
            head.next = next_head
            if next_head:
                copy_node.next = next_head.next
            else:
                copy_node.next = None
            head = next_head
        return res