面试35题:

题目:复杂链表的复制

题: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)

解题思路一:“Python作弊法”

解题代码:

# -*- 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
import copy
return copy.deepcopy(pHead)

解题思路二:分解法。详见剑指offer P188

解题代码:

# -*- 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
self.CloneNodes(pHead)
self.ConnectRandomNodes(pHead)
return self.ReconnectNodes(pHead) def CloneNodes(self,pHead):
'''
复制原始链表的每个结点, 将复制的结点链接在其原始结点的后面
'''
pNode=pHead
while pNode:
pCloned=RandomListNode(0)
pCloned.label=pNode.label
pCloned.next=pNode.next pNode.next=pCloned
pNode=pCloned.next def ConnectRandomNodes(self,pHead):
'''
将复制后的链表中的克隆结点的random指针链接到被克隆结点random指针的后一个结点
'''
pNode=pHead
while pNode:
pCloned=pNode.next
if pNode.random!=None:
pCloned.random=pNode.random.next
pNode=pCloned.next def ReconnectNodes(self,pHead):
'''
拆分链表:将原始链表的结点组成新的链表, 复制结点组成复制后的链表
'''
pNode=pHead
pClonedHead=pClonedNode=pNode.next
pNode.next = pClonedNode.next
pNode=pNode.next
while pNode:
pClonedNode.next=pNode.next
pClonedNode=pClonedNode.next
pNode.next=pClonedNode.next
pNode=pNode.next
return pClonedHead

解法三:递归法,强烈推荐。

# -*- 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
newNode=RandomListNode(pHead.label)
newNode.random=pHead.random
newNode.next=self.Clone(pHead.next)
return newNode
05-04 12:50