19. 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.

 

Example 1:

Input: head = [1,2,3,4,5], n = 2
Output: [1,2,3,5]

Example 2:

Input: head = [1], n = 1
Output: []

Example 3:

Input: head = [1,2], n = 1
Output: [1]

 

Constraints:

  • The number of nodes in the list is sz.
  • 1 <= sz <= 30
  • 0 <= Node.val <= 100
  • 1 <= n <= sz

 

Follow up: Could you do this in one pass?


Solution:

The question is not so hard. We just need to take care of some cases.

1. What if n == length of the given list?
2. If the length of LL is 1
3. What if the head is NULL.
4. General case if the Nth node is between the LL.

See the code below for better understanding.
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* removeNthFromEnd(ListNode* head, int n) {
if((n==1 && head==NULL) || (head->next==NULL && n==1))return NULL;
ListNode* fast=head;
ListNode* slow=head;
while(n--){
fast=fast->next;
}
if(fast==NULL)return head->next;
while(fast->next!=NULL){
fast=fast->next;
slow=slow->next;
}
slow->next=slow->next->next;
return head;
}
};
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