[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。

 

第一步，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。

第二步，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):
        #http://www.cnblogs.com/zuoyuan/p/3745126.html
        if head == None: return None
        # Step 1: duplicate nodes in order
        tmp = head
        while tmp:
            newNode = RandomListNode(tmp.label)
            newNode.next = tmp.next
            tmp.next = newNode
            tmp = tmp.next.next
        # Step 2: preseve random pointer
        tmp = head
        while tmp:
            if tmp.random:
                tmp.next.random = tmp.random.next  # assign the old node's random pointer to new duplicated node's random filed
            tmp = tmp.next.next
        # Step 3: extract new list and return
        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

参考：

http://www.cnblogs.com/zuoyuan/p/3745126.html