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.
复制复杂链表，每个节点有一个random属性随机指向一个任意节点
解法一：
遍历原链表，复制random以外的所有属性，建立原链表和复制链表每个节点的映射，第二部根据字典映射设置每个节点的random
注意映射的字典中要有None:None，不然无法AC
看完评论区别人的代码，再看自己写的代码完全不忍直视。。。。。。。。。。。。。。。。。。。。
1 class Solution(object): 2 def copyRandomList(self, head): 3 """ 4 :type head: RandomListNode 5 :rtype: RandomListNode 6 """ 7 tmphead = head 8 pre = None 9 d = {} 10 d[None]=None 11 res = None 12 while tmphead: 13 tmp = RandomListNode(tmphead.label) 14 d[tmphead] = tmp 15 if pre == None: 16 pre = res = tmp 17 else: 18 pre.next = tmp 19 pre = tmp 20 tmphead = tmphead.next 21 tmphead = head 22 tmpres = res 23 while tmphead: 24 tmpres.random = d[tmphead.random] 25 tmpres = tmpres.next 26 tmphead = tmphead.next 27 return res 28
解法二：
1.在每个节点后面加入一个一样的节点
2.设置每个节点的random
3.分离两个链表
下面的代码有bug，但是没找出来，先放这吧
1 class Solution(object): 2 def copyRandomList(self, head): 3 """ 4 :type head: RandomListNode 5 :rtype: RandomListNode 6 """ 7 tmphead = head 8 while tmphead: 9 tmp = RandomListNode(tmphead.label) 10 node = tmphead.next 11 tmphead.next = tmp 12 tmp.next = node 13 tmphead = node 14 tmphead = head 15 while tmphead: 16 if tmphead.random: 17 tmphead.next.random = tmphead.random.next 18 tmphead = tmphead.next.next 19 tmphead = head 20 res = None 21 ptr = None 22 while tmphead: 23 if res == None: 24 res = tmphead.next 25 ptr = res 26 else: 27 ptr.next = tmphead.next 28 tmphead.next=ptr.next 29 tmphead = tmphead.next 30 return res