Remove Nth Node From End of List Solution
Remove Nth Node From End Of List:
Given the head of a linked list, remove the nth node from the end of the list and return its head.
Two Sum Leetcode Solution :
Constraints :
- The number of nodes in the list is sz.
- 1 <= sz <= 30
- 0 <= Node.val <= 100
- 1 <= n <= sz
Example 1:
Input: head = [1], n = 1
Output: [ ]
Example 2:
Input: head = [1,2], n = 1
Output: [1]
In the above problem, you are given the head of the linked list. The goal is to remove nth node from the end of the list and return it’s head.
Constraints:
- The number of nodes in the list is sz.
- 1 <= sz <= 30
- 0 <= Node.val <= 100
- 1 <= n <= sz
Approach :
Using Two Pointer’s Approach. (slow & fast)
first make the distance between slow and fast by moving fast,while(n–){
fast=fast->next;}then, move slow and fast pointer one by one until
while(fast->next != NULL){
slow=slow->next;
fast=fast->next;
}and in last :-
slow->next = slow->next->next;
return head;but there is an edge case:
if my n = n.size() , in this my fast will point out to NULL
so in we return slow->next;
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Code :
class Solution {
public:
ListNode* removeNthFromEnd(ListNode* head, int n) {
ListNode *slow=head,*fast=head;
while(n--){
fast=fast->next;
}
if(fast==NULL){
return slow->next;
}
while(fast->next != NULL){
slow=slow->next;
fast=fast->next;
}
slow->next = slow->next->next;
return head;
}
};
class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
if(head == null) return head;
int k = 0;
ListNode curr = head;
for(; curr != null; curr = curr.next) k++;
k -= n; //to get position of given element from first node
if(k == 0) return head.next;
for(curr = head; k > 1; k--, curr = curr.next) ;
curr.next = curr.next.next; // to remove that node
return head;
}
}
class Solution:
def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
fast, slow = head, head
for _ in range(n): fast = fast.next
if not fast: return head.next
while fast.next: fast, slow = fast.next, slow.next
slow.next = slow.next.next
return head
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