Remove Nth Node From End of List

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.

Solution:
两个指针fast和slow，fast先走n步，然后两个再一起走，当fast走到最后一个的时候，slow指的就是倒数第n个节点。

Run time complexity is O(n), constant space.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        if(head == null || head.next == null) {  //if only has one node, after removed, it will be null
            return null;
        }
        
        ListNode fast = head;
        ListNode slow = head;
        while(n != 0) {
            fast = fast.next;
            n--;
        }
        
        if(fast == null) {  //fast==null means the node which need to be removed is the head
            head = head.next;
            return head;
        }
        
        while(fast.next != null) {
            fast = fast.next;
            slow = slow.next;
        }
        
        slow.next = slow.next.next;

        return head;
    }
}