In this Delete duplicate-value nodes from a sorted linked list HackerRank solution, You are given the pointer to the head node of a sorted linked list, where the data in the nodes is in ascending order. Delete nodes and return a sorted list with each distinct value in the original list. The given head pointer may be null indicating that the list is empty.
Example
refers to the first node in the list .
Remove 1 of the data values and return pointing to the revised list .
Function Description
Complete the removeDuplicates function in the editor below.
removeDuplicates has the following parameter:
- SinglyLinkedListNode pointer head: a reference to the head of the list
Returns
- SinglyLinkedListNode pointer: a reference to the head of the revised list
Input Format
The first line contains an integer , the number of test cases.
The format for each test case is as follows:
The first line contains an integer , the number of elements in the linked list.
Each of the next lines contains an integer, the value for each of the elements of the linked list.
Constraints
Sample Input
STDIN Function
----- --------
1 t = 1
5 n = 5
1 data values = 1, 2, 2, 3, 4
2
2
3
4
Sample Output
1 2 3 4
Explanation
The initial linked list is: .
The final linked list is: .
Delete duplicate-value nodes from a sorted 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 C
SinglyLinkedListNode* removeDuplicates(SinglyLinkedListNode* llist) {
SinglyLinkedListNode* temp=llist;
SinglyLinkedListNode* temp2=NULL;
while(temp->next!=NULL){
if(temp->data==temp->next->data){
temp2=temp->next->next;
free(temp->next);
temp->next=temp2;
}
else{
temp=temp->next;
}
}
return llist;
}Problem Solution in C++
SinglyLinkedListNode* removeDuplicates(SinglyLinkedListNode* llist) {
SinglyLinkedListNode* prev=llist;
SinglyLinkedListNode* curr=llist->next;
SinglyLinkedListNode* temp;
while(curr!=nullptr){
if(prev->data==curr->data){
prev->next=curr->next;
temp=curr;
curr=curr->next;
free(temp);
}
else{
prev=prev->next;
curr=curr->next;
}
}
return llist;
}Problem Solution in Python
def removeDuplicates(llist):
if llist == None:
return
current = llist
while(current.next):
if current.data == current.next.data:
current.next = current.next.next
else:
current = current.next
return llistProblem Solution in JavaScript
function removeDuplicates(llist) {
let current = llist;
if(llist === null)
return;
while(current !== null){
console.log(current.data);
if(current.next === null)
break;
if(current.data === current.next.data){
current.next = current.next.next;
}
else
current = current.next;
}
return llist;
}