Delete the Nth node from the end of the LinkedList

//tc:O(N) n is the length of the linked list //sc :(1) constant /** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode() {} * ListNode(int val) { this.val = val; } * ListNode(int val, ListNode next) { this.val = val; this.next = next; } * } */ class Solution { public ListNode removeNthFromEnd(ListNode head, int n) { // base case if(head.next ==null && n ==1) return null; ListNode node1 = head; ListNode node2 = head; //prev is required for edge case example if the node to be deleted is the last node of the list from start ListNode prev = new ListNode(0); prev.next = head; // let node1 go ahead of node2 by n nodes while(n-->0){ node1 = node1.next; } // when node1 will reach the end node then node2 will have reached the node to be delete while(node1!=null){ node1 = node1.next; node2 = node2.next; prev = prev.next; } //here is the edge case what if the node to be deleted is the last node if(node2.next ==null){ prev.next = node2.next; // prev of node2 will come in handy } else{ node2.val = node2.next.val; node2.next = node2.next.next; } return head; } } 

/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode() {} * ListNode(int val) { this.val = val; } * ListNode(int val, ListNode next) { this.val = val; this.next = next; } * } */ class Solution { public ListNode removeNthFromEnd(ListNode head, int n) { //without using prev pointer ListNode faster = head; ListNode slower = head; while(n-->0){ faster = faster.next; } //if faster is null then the node to be delete is the head node if(faster ==null) { return head.next; } while(faster.next!=null){ faster = faster.next; slower = slower.next; } slower.next = slower.next.next; return head; } }