Given a linked list, remove the nth node from the end of list and return its head.
For 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. Try to do this in one pass.
/**
* Definition for singly-linked list.
* function ListNode(val) {
* this.val = val;
* this.next = null;
* }
*/
/**
* @param {ListNode} head
* @param {number} n
* @return {ListNode}
*/
var removeNthFromEnd = function(head, n) {
var l = 0;
var pointer = head;
while (pointer !== null) {
l++;
pointer = pointer.next;
}
var index = l - n;
pointer = head;
if (index === 0) {
return pointer.next;
}
for (var i = 0; i < index; i++) {
if (i === index - 1) {
pointer.next = pointer.next.next;
} else {
pointer = pointer.next;
}
}
return head;
};