DevExCode

19. Remove Nth Node From End of List

Linked ListTwo Pointers

Explanation

To remove the nth node from the end of a linked list, we can use a two-pointer approach. We initialize two pointers, fast and slow, both pointing to the head of the linked list. We move the fast pointer to the nth node from the beginning. Then, we move both pointers simultaneously until the fast pointer reaches the end of the list. At this point, the slow pointer will be pointing to the node just before the node to be removed. We update the slow pointer to skip the nth node from the end, effectively removing it from the list.

View Leetcode Problem
class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode fast = dummy;
        ListNode slow = dummy;

        for (int i = 0; i <= n; i++) {
            fast = fast.next;
        }

        while (fast != null) {
            fast = fast.next;
            slow = slow.next;
        }

        slow.next = slow.next.next;

        return dummy.next;
    }
}

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