[leetcode]Copy List with Random Pointer @ Python

原题地址：https://oj.leetcode.com/problems/copy-list-with-random-pointer/

题意：

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.

解题思路：这题主要是需要深拷贝。看图就明白怎么写程序了。

 

首先，在原链表的每个节点后面都插入一个新节点，新节点的内容和前面的节点一样。比如上图，1后面插入1，2后面插入2，依次类推。

其次，原链表中的random指针如何映射呢？比如上图中，1节点的random指针指向3，4节点的random指针指向2。如果有一个tmp指针指向1（蓝色），则一条语句：tmp.next.random = tmp.random.next；就可以解决这个问题。

第三步，将新的链表从上图这样的链表中拆分出来。

代码：

# Definition for singly-linked list with a random pointer.

# class RandomListNode:

#     def __init__(self, x):

#         self.label = x

#         self.next = None

#         self.random = None



class Solution:

    # @param head, a RandomListNode

    # @return a RandomListNode

    def copyRandomList(self, head):

        if head == None: return None

        tmp = head

        while tmp:

            newNode = RandomListNode(tmp.label)

            newNode.next = tmp.next

            tmp.next = newNode

            tmp = tmp.next.next

        tmp = head

        while tmp:

            if tmp.random:

                tmp.next.random = tmp.random.next

            tmp = tmp.next.next

        newhead = head.next

        pold = head

        pnew = newhead

        while pnew.next:

            pold.next = pnew.next

            pold = pold.next

            pnew.next = pold.next

            pnew = pnew.next

        pold.next = None

        pnew.next = None

        return newhead