Javascript Program For Cloning A Linked List With Next And Random Pointer In O(1) Space
Last Updated :
22 Jun, 2022
Given a linked list having two pointers in each node. The first one points to the next node of the list, however, the other pointer is random and can point to any node of the list. Write a program that clones the given list in O(1) space, i.e., without any extra space.
Examples:
Input : Head of the below-linked list
Output :
A new linked list identical to the original list.
In the previous posts Set-1 and Set-2 various methods are discussed, and O(n) space complexity implementation is also available.
In this post, we’ll be implementing an algorithm that’d require no additional space as discussed in Set-1.
Below is the Algorithm:
- Create the copy of node 1 and insert it between node 1 & node 2 in the original Linked List, create a copy of 2 and insert it between 2 & 3. Continue in this fashion, add the copy of N after the Nth node
- Now copy the random link in this fashion
original->next->random= original->random->next; /*TRAVERSE
TWO NODES*/
- This works because original->next is nothing but a copy of the original and Original->random->next is nothing but a copy of the random.
- Now restore the original and copy linked lists in this fashion in a single loop.
original->next = original->next->next;
copy->next = copy->next->next;
- Ensure that original->next is NULL and return the cloned list
Below is the implementation.
Javascript
<script>
class Node {
constructor(x) {
this .data = x;
this .next = this .random = null ;
}
}
function print(start) {
var ptr = start;
while (ptr != null ) {
document.write(
"Data = " +
ptr.data + ", Random = "
+ ptr.random.data+ "<br/>"
);
ptr = ptr.next;
}
}
function clone(start) {
var curr = start, temp = null ;
while (curr != null ) {
temp = curr.next;
curr.next = new Node(curr.data);
curr.next.next = temp;
curr = temp;
}
curr = start;
while (curr != null ) {
if (curr.next != null )
curr.next.random = (curr.random != null ) ?
curr.random.next : curr.random;
curr = (curr.next != null ) ?
curr.next.next : curr.next;
}
var original = start, copy = start.next;
temp = copy;
while (original != null && copy != null ) {
original.next = (original.next != null ) ?
original.next.next : original.next;
copy.next = (copy.next != null ) ?
copy.next.next : copy.next;
original = original.next;
copy = copy.next;
}
return temp;
}
var start = new Node(1);
start.next = new Node(2);
start.next.next = new Node(3);
start.next.next.next = new Node(4);
start.next.next.next.next = new Node(5);
start.random = start.next.next;
start.next.random = start;
start.next.next.random =
start.next.next.next.next;
start.next.next.next.random =
start.next.next.next.next;
start.next.next.next.next.random =
start.next;
document.write( "Original list : <br/>" );
print(start);
document.write( "<br>" );
document.write( "Cloned list : <br/>" );
var cloned_list = clone(start);
print(cloned_list);
</script>
|
Output
Original list :
Data = 1, Random = 3
Data = 2, Random = 1
Data = 3, Random = 5
Data = 4, Random = 5
Data = 5, Random = 2
Cloned list :
Data = 1, Random = 3
Data = 2, Random = 1
Data = 3, Random = 5
Data = 4, Random = 5
Data = 5, Random = 2
Time Complexity: O(n), where n is the number of nodes in the given linked list.
Auxiliary Space: O(1), as no extra space is used. The n nodes which are inserted in between the nodes was already required to clone the list, so we can say that we did not use any extra space.
Please refer complete article on Clone a linked list with next and random pointer in O(1) space for more details!
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