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)
第一反应:分两步。第一步考虑next,第二步考虑random。利用哈希表建立映射关系。
# -*- 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:
p1 = pHead
p2 = RandomListNode(p1.label)
pHeadNew = p2
hashMap = {}
while p1.next:
hashMap[p1] = p2
p2.next = RandomListNode(p1.next.label)
p1 = p1.next
p2 = p2.next
hashMap[p1] = p2
p1 = pHead
p2 = pHeadNew
while p1:
if p1.random in hashMap:
p2.random = hashMap[p1.random]
p1 = p1.next
p2 = p2.next
return pHeadNew
else:
return None
进阶版:将复制的每个结点接在对应原始结点的后面,复制结点random指针与原始结点random指针指向的位置应该是相邻的,最后将链表拆为原始链表和复制链表。
# -*- 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:
p1 = pHead
while p1:
p2 = RandomListNode(p1.label)
p2.next = p1.next
p1.next = p2
p1 = p2.next
p1 = pHead
while p1:
p2 = p1.next
if p1.random:
p2.random = p1.random.next
p1 = p2.next
p1 = pHead
pHeadNew = p2 = p1.next
p1.next = p2.next
p1 = p1.next
while p1:
p2.next = p1.next
p2 = p2.next
p1.next = p2.next
p1 = p1.next
return pHeadNew
else:
return None