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.

一刷
题解： 快慢指针
Time Complexity - O(n)， Space Complexity - O(1)。

public class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode fast = head;
        ListNode slow = head;
        while(n>0){
            if(fast == null) return null;
            fast = fast.next;
            n--;
        }
        //when fast == null, slow is to remove
        if(fast == null){
            head = head.next;
            return head;
        }
        
        while(fast.next!=null){
            fast = fast.next;
            slow = slow.next;
        }
        
        //slow next is to remove
        slow.next = slow.next.next;
        return head;
    }
}

二刷：
如何避免空指针的情况：用dummy node， 所有节点从dummy node开始。第二，利用快慢指针，快指针先往前移动n+1个位置。
然后快慢都往前移动直到快指针移动到null(避免调用fast.next). 那么slow的下一位的就是要被移除的点（slow一定不为null）

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode start = new ListNode(0);
        ListNode slow = start, fast = start;
        slow.next = head;
        
        while(n>=0){
            fast = fast.next;
            n--;
        }
        
        while(fast!=null){
            fast = fast.next;
            slow = slow.next;
        }
        
        slow.next = slow.next.next;
        return start.next;
    }
}