138. Copy List with Random Pointer

https://leetcode.com/problems/copy-list-with-random-pointer/#/description

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.

Sol 1:

# Definition for singly-linked list with a random pointer.
# class RandomListNode(object):
#     def __init__(self, x):
#         self.label = x
#         self.next = None
#         self.random = None

class Solution(object):
    def copyRandomList(self, head):
        """
        :type head: RandomListNode
        :rtype: RandomListNode
        """
        # Time O（2n） Space O(1) 

        dic = dict()
        m = n = head
        
        # m is to copy label data structure, one pass. Time O(n) 
        while m:
            dic[m] = RandomListNode(m.label)
            m = m.next
        # n is to copy next and random data structure. one pass. Time O(n)
        while n:
            dic[n].next = dic.get(n.next)
            dic[n].random = dic.get(n.random)
            n = n.next
        return dic.get(head)

Note:

1 In the initlization of a linked list, 

self.label = x

"label" here mean value.... it's a name, does not matter

Sol 2:

# Definition for singly-linked list with a random pointer.
# class RandomListNode(object):
#     def __init__(self, x):
#         self.label = x
#         self.next = None
#         self.random = None

import collections
class Solution(object):
    def copyRandomList(self, head):
        """
        :type head: RandomListNode
        :rtype: RandomListNode
        """
        # Time O（n） Space O(1) 

        # Nice lambda initlization here. Pythonic
        dic = collections.defaultdict(lambda : RandomListNode(0))
        dic[None] = None
        
        n = head

        while n:
            dic[n].label = n.label
            dic[n].next = dic[n.next]
            dic[n].random = dic[n.random]
            n = n.next
        return dic.get(head)