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# 138. Copy List with Random Pointer

"""
A linked list of length n is given such that each node contains an additional
random pointer, which could point to any node in the list, or null.

Construct a deep copy of the list. The deep copy should consist of exactly n
brand new nodes, where each new node has its value set to the value of its
corresponding original node. Both the next and random pointer of the new nodes
should point to new nodes in the copied list such that the pointers in the
original list and copied list represent the same list state. None of the
pointers in the new list should point to nodes in the original list.

For example, if there are two nodes X and Y in the original list, where
X.random
--> Y, then for the corresponding two nodes x and y in the copied list,
x.random
--> y.

Return the head of the copied linked list.

The linked list is represented in the input/output as a list of n nodes. Each
node is represented as a pair of [val, random_index] where:

  val: an integer representing Node.val
  random_index: the index of the node (range from 0 to n-1) that the random
pointer points to, or null if it does not point to any node.

Your code will only be given the head of the original linked list.
"""


class Node:
    def __init__(self, x, next=None, random=None):
        self.val = int(x)
        self.next = next
        self.random = random


class Solution:
    def copyRandomList(self, head):
        cache = {}
        new_ptr = curr = Node(-1)
        while head:
            if head in cache:
                newnode = cache[head]
            else:
                newnode = Node(head.val)
                cache[head] = newnode
            new_ptr.next = newnode
            new_ptr = new_ptr.next
            if head.random:
                if head.random in cache:
                    new_random = cache[head.random]
                else:
                    new_random = Node(head.random.val)
                    cache[head.random] = new_random
                new_ptr.random = new_random
            head = head.next
        return curr.next