[leetcode]Copy List with Random Pointer @ Python

原題地址：https://oj.leetcode.com/problems/copy-list-with-random-pointer/

題意：

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.

解題思路：這題主要是需要深拷貝。看圖就明白怎么寫程序了。

 

首先，在原鏈表的每個節點后面都插入一個新節點，新節點的內容和前面的節點一樣。比如上圖，1后面插入1，2后面插入2，依次類推。

其次，原鏈表中的random指針如何映射呢？比如上圖中，1節點的random指針指向3，4節點的random指針指向2。如果有一個tmp指針指向1（藍色），則一條語句：tmp.next.random = tmp.random.next；就可以解決這個問題。

第三步，將新的鏈表從上圖這樣的鏈表中拆分出來。

代碼：

# 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