Copy List With Random Pointer

Problem

https://leetcode.com/problems/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.

Example 1:

Input: head = [[7,null],[13,0],[11,4],[10,2],[1,0]]
Output: [[7,null],[13,0],[11,4],[10,2],[1,0]]

Example 2:

Input: head = [[1,1],[2,1]]
Output: [[1,1],[2,1]]

Example 3:

Input: head = [[3,null],[3,0],[3,null]]
Output: [[3,null],[3,0],[3,null]]

Constraints:

• 0 <= n <= 1000

• -10:sup:`4`<= Node.val <= 10:sup:`4`

• Node.random is null or is pointing to some node in the linked list.

Pattern

Hash Table, Linked List

Approaches

Hash Map

Explanation

The trick is to use a old2copy = defaultdict(Node) to map the original node to its copy. The defaultdict creates the copy automatically if one doesn’t exist which greatly simplifies the code. We add old2copy[None] = None to keep pointers to None in the copy if they were in the original.

We use a standard while node: node = node.next loop to go through all the elements of the original linked list. For each node, we set the value of the copied node to be the same as the original node and

copy.next = old2copy[node.next]
copy.random = old2copy[node.random]

We keep track of the head so that we can return the copy old2copy[head] at the end.

Code

from __future__ import annotations

from collections import defaultdict


class Node:
    """Node in a linked list with a random pointer."""

    def __init__(
        self,
        x: int = 0,
        next: Node | None = None,
        random: Node | None = None,
    ):
        self.val = x
        self.next = next
        self.random = random


def copyRandomList(head: Node | None) -> Node | None:
    """Return a deep copy of the linked list with random pointers."""
    if head is None:
        return None

    old2copy = defaultdict(lambda: Node(0))
    old2copy[None] = None

    node = head
    while node:
        copy = old2copy[node]
        copy.val = node.val
        copy.next = old2copy[node.next]
        copy.random = old2copy[node.random]

        node = node.next

    return old2copy[head]

Test

>>> from copy_list_with_random_pointer__hash_map import copyRandomList, Node
>>> n1, n2, n3 = Node(7), Node(13), Node(11)
>>> n1.next, n2.next = n2, n3
>>> n2.random, n3.random = n1, n1
>>> c = copyRandomList(n1)
>>> [c.val, c.next.val, c.next.next.val]
[7, 13, 11]
>>> c is not n1
True

Complexity

\(n\) is the number of elements in the linked list

Measure	Complexity	Notes
Time	\(O(n)\)	one pass through the linked list
Auxiliary Space	\(O(n)\)	old2copy dictionary stores all the original nodes and copied nodes

class copy_list_with_random_pointer__hash_map.Node(x: int = 0, next: Node | None = None, random: Node | None = None)

Bases: object

Node in a linked list with a random pointer.

copy_list_with_random_pointer__hash_map.copyRandomList(head: Node | None) → Node | None

Return a deep copy of the linked list with random pointers.