Remove Nth Node From end of List

public ListNode removeNthFromEnd(ListNode head, int n) {
if(head == null) return null;
if(n == 0) return head;
if(head.next == null && n==1) return null;
ListNode runner = head;
//ListNode p = new ListNode(0);
//p.next = head;
ListNode p = head;

int count = 1;
while(count<=n && runner.next!=null){
count++;
runner = runner.next;
}

if(count == n) return head.next;

while(runner.next!=null){
p = p.next;
runner = runner.next;
}
// runner points to the end of list, p points to the previous one of target node

p.next = p.next.next;
//p = p.next;
return head;
}


Question:

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.
Answer:
in one pass，所以基本思路是一快一慢两个指针