# Edlison

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# 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.

Solution One

``````class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
int len = 1;

if (p == null)
return null;

while (p.next != null) {
len++;
p = p.next;
}

int loc = len - n;

if (loc == 0) {
}

for (int i = 1; i < loc; i++) {
temp = temp.next;
}

temp.next = temp.next.next;

}
}
``````

Pass one time and get the length of LinkedList, so just count `length - n - 1` times we can find the pre pointer. All we need to do is point pre pointer.next equal the next of which pointer we should delete.

Solution Two

``````class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {

while (temp != null) {
temp = temp.next;
}

ListNode[] list = stack.toArray(new ListNode[stack.size()]);

if ((list.length - n - 1) < 0) return head.next;

list[list.length - n - 1].next = list[list.length - n].next;

Just let the List structure give a help, we can transfer a `LinkedList` to an `Array`. So all we need to do is let `list[list.length - n - 1].next` equals `list[list.length - n].next`.
Base case: if delete the head node we just need to return `head.next`.