Medium
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.valrandom_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]]
Example 4:
Input: head = []
Output: []
Explanation: The given linked list is empty (null pointer), so return null.
Constraints:
0 <= n <= 1000-10000 <= Node.val <= 10000Node.random is null or is pointing to some node in the linked list.To solve this problem with the Solution class, we can use a hashmap to keep track of the mapping between original nodes and their corresponding copied nodes. Here are the steps:
Solution class with a method copyRandomList that takes the head of the original linked list as input and returns the head of the copied linked list.copyRandomList method, if the input head is None, return None as the copied list is empty.next and random pointers of the copied node to None.next and random pointers of copied nodes according to the mapping stored in the hashmap.Here’s how the Solution class would look like in Python:
class Solution:
def copyRandomList(self, head: 'Node') -> 'Node':
if not head:
return None
mapping = {}
# Create copies of nodes without pointers
current = head
while current:
mapping[current] = Node(current.val)
current = current.next
# Assign pointers for each copied node
current = head
while current:
mapping[current].next = mapping.get(current.next)
mapping[current].random = mapping.get(current.random)
current = current.next
return mapping[head]
Make sure to define the Node class before using it in the copyRandomList method. This solution constructs a deep copy of the original linked list while maintaining the relationships between nodes as specified.
class Node:
def __init__(self, x: int, next: 'Node' = None, random: 'Node' = None):
self.val = int(x)
self.next = next
self.random = random
"""
# Definition for a Node.
class Node:
def __init__(self, x: int, next: 'Node' = None, random: 'Node' = None):
self.val = int(x)
self.next = next
self.random = random
"""
class Solution:
def copyRandomList(self, head: 'Optional[Node]') -> 'Optional[Node]':
if head is None:
return None
# First pass: create the cloned nodes and insert them right after the original nodes
curr = head
while curr:
cloned_node = Node(curr.val)
cloned_node.next = curr.next
curr.next = cloned_node
curr = cloned_node.next
# Second pass: update the random pointers of the cloned nodes
curr = head
while curr:
if curr.random:
curr.next.random = curr.random.next
curr = curr.next.next
# Third pass: separate the cloned list from the original list
curr = head
new_head = head.next
while curr:
cloned_node = curr.next
curr.next = cloned_node.next
if cloned_node.next:
cloned_node.next = cloned_node.next.next
curr = curr.next
return new_head