Last updated on July 19th, 2024 at 07:36 pm

Here, We see Copy List with Random Pointer LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

List of all LeetCode Solution

Topics

Hash Table, Linked List

Companies

Amazon, Bloomberg, Microsoft, Uber

Level of Question

Medium

Copy List with Random Pointer LeetCode Solution

Problem Statement

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]]

1. Copy List with Random Pointer LeetCode Solution C++

class Solution {
public:
    Node* copyRandomList(Node* head) {
        if (!head) return nullptr;
        unordered_map<Node*, Node*> old_to_new;
        Node* curr = head;
        while (curr) {
            old_to_new[curr] = new Node(curr->val);
            curr = curr->next;
        }
        curr = head;
        while (curr) {
            old_to_new[curr]->next = old_to_new[curr->next];
            old_to_new[curr]->random = old_to_new[curr->random];
            curr = curr->next;
        }
        return old_to_new[head];
    }
};

2. Copy List with Random Pointer LeetCode Solution Java

public class Solution {
    public Node copyRandomList(Node head) {
        if (head == null) return null;
        HashMap<Node, Node> oldToNew = new HashMap<>();
        Node curr = head;
        while (curr != null) {
            oldToNew.put(curr, new Node(curr.val));
            curr = curr.next;
        }
        curr = head;
        while (curr != null) {
            oldToNew.get(curr).next = oldToNew.get(curr.next);
            oldToNew.get(curr).random = oldToNew.get(curr.random);
            curr = curr.next;
        }
        return oldToNew.get(head);
    }
}

3. Copy List with Random Pointer LeetCode Solution JavaScript

var copyRandomList = function(head) {
    if (!head) return null;
    const oldToNew = new Map();
    let curr = head;
    while (curr) {
        oldToNew.set(curr, new Node(curr.val));
        curr = curr.next;
    }
    curr = head;
    while (curr) {
        oldToNew.get(curr).next = oldToNew.get(curr.next) || null;
        oldToNew.get(curr).random = oldToNew.get(curr.random) || null;
        curr = curr.next;
    }
    return oldToNew.get(head);
};

4. Copy List with Random Pointer Solution Python

class Solution(object):
    def copyRandomList(self, head):
        if not head:
            return None
        old_to_new = {}
        curr = head
        while curr:
            old_to_new[curr] = Node(curr.val)
            curr = curr.next
        curr = head
        while curr:
            old_to_new[curr].next = old_to_new.get(curr.next)
            old_to_new[curr].random = old_to_new.get(curr.random)
            curr = curr.next
        return old_to_new[head]