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