leetcode:复杂链表的复制

原题：https://leetcode-cn.com/problems/fu-za-lian-biao-de-fu-zhi-lcof/

我的解法

# Definition for a Node.
class Node:
    def __init__(self, x, next=None, random=None):
        self.val = int(x)
        self.next = next
        self.random = random

class Solution(object):
    def copyRandomList(self, head):
        """
        :type head: Node
        :rtype: Node
        """
        if head==None:
            return None
        hd=head
        thead=Node(hd.val)
        t=thead
        list2=[hd.random]
        hd=hd.next
        list=[thead]
        while hd!=None:
            n=Node(hd.val)
            t.next=n
            list2.append(hd.random)
            list.append(n)
            t=n
            hd=hd.next
        t1=thead
        for i in range(len(list2)):
            if list2[i]==None:
                t1.random=None
                t1=t1.next
                continue
            j=0
            hd=head
            while hd!=None:
                if hd==list2[i]:
                    t1.random=list[j]
                    t1=t1.next
                    break
                j+=1
                hd=hd.next
        return thead

官方解法

使用哈希和回溯法。

哈希 键：原链表结点    值：复制结点

回溯：让每个结点的拷贝相互独立

# Definition for a Node.
class Node:
    def __init__(self, x, next=None, random=None):
        self.val = int(x)
        self.next = next
        self.random = random

class Solution(object):
    def __init__(self):
        self.hashNode={}
    def copyRandomList(self, head):
        if head==None:
            return None
        if head not in self.hashNode.keys():
            headNew=Node(head.val)
            self.hashNode[head]=headNew
            headNew.next=self.copyRandomList(head.next)
            headNew.random = self.copyRandomList(head.random)
        return self.hashNode[head]