## DEV Community is a community of 892,765 amazing developers

We're a place where coders share, stay up-to-date and grow their careers.

seanpgallivan

Posted on

# Solution: Copy List with Random Pointer

This is part of a series of Leetcode solution explanations (index). If you liked this solution or found it useful, please like this post and/or upvote my solution post on Leetcode's forums.

#### Leetcode Problem #138 (Medium): Copy List with Random Pointer

Description:

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`.

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.

Examples:

Example 1:
Output: [[7,null],[13,0],[11,4],[10,2],[1,0]]
Visual:
Example 2:
Output: [[1,1],[2,1]]
Visual:
Example 3:
Output: [[3,null],[3,0],[3,null]]
Visual:
Example 4:
Output: []
Explanation: The given linked list is empty (null pointer), so return null.

Constraints:

• `0 <= n <= 1000`
• `-10000 <= Node.val <= 10000`
• `Node.random` is `null` or is pointing to some node in the linked list.

Idea:

The tricky thing here is that the nodes have nothing uniquely identifying, except for their pointer, since there are no indexes and even the values can be duplicates. At this point, the easiest thing to do would seem to be to just find a way to reference together each original nodes and its copy.

For that, we can use a pointer map. The pointer map will simply be a reference/lookup between the two nodes, so we'll use the original node pointer as the key and the new node pointer as the value in our key/value pairs.

Since the random nodes can be anywhere in the linked list, even past the current node, we'll go ahead and run through the list twice. The first time through, we'll just create the new nodes, form our standard link through the .next attribute, and store the reference in pmap.

Then we can make our second pass and use the references in pmap to properly assign the .random attribute for each node.

Javascript Code:

``````var copyRandomList = function(head) {
let pmap = new Map(), dummy = {},
curr = head, copy = dummy
while (curr) {
let newNode = new Node(curr.val, null, null)
pmap.set(curr, newNode)
copy.next = newNode, copy = newNode, curr = curr.next
}
curr = head, copy = dummy.next
while (curr) {
copy.random = pmap.get(curr.random)
curr = curr.next, copy = copy.next
}
return dummy.next
};
``````

## Discussion (1)

AhmedSiam72

It is a best solution found that very popular and helpful: