【LeetCode with Python】 Copy List with Random Pointer

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原题页面：https://oj.leetcode.com/problems/copy-list-with-random-pointer/
题目类型：
难度评价：★
本文地址：http://blog.csdn.net/nerv3x3/article/details/39453281

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.

# Definition for singly-linked list with a random pointer.class RandomListNode:    def __init__(self, x):        self.label = x        self.next = None        self.random = Noneclass Solution:    # @param head, a RandomListNode    # @return a RandomListNode    def copyRandomList(self, head):        if None == head:            return None        save_list = [ ]        p1 = head        while None != p1:            save_list.append(p1)            p1 = p1.next        new_head = RandomListNode(-1)        new_head.next = head        first = new_head        second = head        copy_head = RandomListNode(-1)        copy_first = copy_head        copy_second = None        while None != first:            copy_second = RandomListNode(second.label) if None != second else None            copy_first.next = copy_second            copy_first = copy_first.next            first = first.next            if None != second:                second = second.next        p1 = head        p1_tail = head.next        p2 = copy_head.next        while None != p1:            p1_tail = p1.next            p1.next = p2            p2.random = p1            p1 = p1_tail            p2 = p2.next        p2 = copy_head.next        #p1 = head        while None != p2:            p2.random = p2.random.random.next if None != p2.random.random else None            #p1.next = p1.next.next.random if None != p1.next.next else None   # may broken the previous p1.next, so have to save ori list            p2 = p2.next            #p1 = p1.next        len_save_list = len(save_list)        for i in range(0, len_save_list - 1):            save_list[i].next = save_list[i + 1]        save_list[len_save_list - 1].next = None        return copy_head.next