Delete a node| Linked List
Task (easy)
Delete the node at a given position in a linked list and return a reference to the head node. The head is at position 0. The list may be empty after you delete the node. In that case, return a null value.
Example
Llist=0->1->2->3
position=2
After removing the node at position 2, llist’= 0->1->3.
Function Description
Complete the deleteNode function in the editor below. deleteNode has the following parameters:
- SinglyLinkedListNode pointer llist: a reference to the head node in the list
- int position: the position of the node to remove
Returns
- SinglyLinkedListNode pointer: a reference to the head of the modified list
Input Format
The first line of input contains an integer n, the number of elements in the linked list.
Each of the next n lines contains an integer, the node data values in order.
The last line contains an integer, position, the position of the node to delete.
Constraints
1<n<1000
1<list[i]<1000, where list[i] is the ith element of the linked list.
Sample Input
8
20
6
2
19
7
4
15
9
3
Sample Output
20 6 2 7 4 15 9
Solution 1
def deleteNode(llist, position):
if position == 0:
return llist.next
node = llist
for _ in range(position - 1):
node = node.next
node.next = node.next.next
return llist
Solution 2
def deleteNode(llist, position):
if position == 0:
return llist.next
i = 0
cur = llist
while cur:
if position-1 == i:
break
i+=1
cur = cur.next
cur.next = cur.next.next
return llist
Explanation Steps
1.Check if the list is empty.
2. Handle deletion of the head node.
3.Traverse the list to find the node to delete.
4. Update pointers to remove the node from the list.
