In this Reverse a linked list HackerRank solution, Given the pointer to the head node of a linked list, change the next pointers of the nodes so that their order is reversed. The head pointer given may be null meaning that the initial list is empty.
Example
references the list
Manipulate the pointers of each node in place and return , now referencing the head of the list .
Function Description
Complete the reverse function in the editor below.
reverse has the following parameter:
- SinglyLinkedListNode pointer head: a reference to the head of a list
Returns
- SinglyLinkedListNode pointer: a reference to the head of the reversed list
Input Format
The first line contains an integer , the number of test cases.
Each test case has the following format:
The first line contains an integer , the number of elements in the linked list.
Each of the next lines contains an integer, the values of the elements in the linked list.
Constraints
- , where is the element in the list.
Sample Input
1
5
1
2
3
4
5
Sample Output
5 4 3 2 1
Explanation
The initial linked list is: .
The reversed linked list is: .
Reverse a linked list HackerRank solution
I will Provide solution in Multiple programming languages for you. If you are not able to find the code in required language then please share in comments so that our team can help you.
Problem Solution in Python
def reverse(llist):
# Write your code here
prev = None
while llist != None:
# get next
nxt = llist.next
# point current to prev
llist.next = prev
# update prev before next iteration
prev = llist
# update head to be nxt and restart
llist = nxt
return prevProblem Solution in Java
public static SinglyLinkedListNode reverse(SinglyLinkedListNode llist) {
//head = llist
if (llist == null) {
return null;
}
//single node
if (llist.next == null) {
return llist;
}
SinglyLinkedListNode prevNode = null;
SinglyLinkedListNode currNode = llist;
while (currNode != null) {
SinglyLinkedListNode nextNode = currNode.next;
currNode.next = prevNode;
prevNode = currNode;
currNode = nextNode;
}
llist = prevNode;
return llist;
}Problem Solution in JavaScript
const reverse = head => {
let prev = null
while (head) {
const next = head.next
head.next = prev
prev = head
head = next
}
return prev
}Problem Solution in C++
SinglyLinkedListNode* reverse(SinglyLinkedListNode* llist) {
SinglyLinkedListNode* prev = nullptr;
SinglyLinkedListNode* next = llist;
while(llist && next){
next = llist->next;
llist->next = prev;
prev = llist;
if(next)
llist = next;
}
return llist;
}