Remove Nth Node From End of List

Given a linked list, remove the n-th node from the end of list and return its head.

Example:
Given linked list: 1->2->3->4->5 and n=2.

After removing the second node from the end, the linked list becomes 1->2->3->5.

Note: Given n will always be valid.

Follow up:
could you do this in one pass?

Solution for this problem statement:

public ListNode removeNthFromEnd(ListNode head, int n) {
    ListNode dummy = new ListNode(0);
    dummy.next = head;
    ListNode first = dummy;
    ListNode second = dummy;
    // Advances first pointer so that the gap between first and second is n nodes apart
    for (int i = 1; i <= n + 1; i++) {
        first = first.next;
    }
    // Move first to the end, maintaining the gap
    while (first != null) {
        first = first.next;
        second = second.next;
    }
    second.next = second.next.next;
    return dummy.next;
}

There is another solution is available for this problem statement:

public ListNode removeNthFromEnd(ListNode head, int n) {
    ListNode dummy = new ListNode(0);
    dummy.next = head;
    int length  = 0;
    ListNode first = head;
    while (first != null) {
        length++;
        first = first.next;
    }
    length -= n;
    first = dummy;
    while (length > 0) {
        length--;
        first = first.next;
    }
    first.next = first.next.next;
    return dummy.next;
}

There are other articles are available which give you detailed information about the array. Also, there are some linked list-related programs are available like Add two numbers using a linked list.

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