LeetCode刷题笔录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.

用两个指针，s和f。让f先前进n步，这时候有两种情况：

1.f=null，那说明n就等于这个链表的长度，因此要删除的是头指针。return head.next

2.f!=null。这时将f和s以同样的速度前移，直到f到达链表的最后一位，即f.next=null。此时s的位置处于要被删除的node的前一位，即倒数第n+1为。s.next=s.next.next即可。

/**
 * 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) {
        ListNode s = head;
        ListNode f = head;
        for(int i = 1; i <= n; i++){
            f = f.next;
        }
        //if f == null, then the first node is to be deleted
        if(f == null)
            return head.next;
        else{
            while(f.next != null){
                f = f.next;
                s = s.next;
            }
        s.next = s.next.next;
        }
        
        return head;
    }
}