[leetcode]19. Remove Nth Node From End of List删除链表倒数第N个节点

Given a linked list, remove the n-th node from the end of list and return its head.

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.

Follow up:

Could you do this in one pass?

题意： 删除链表倒数第N个节点

Solution1: Two Pointers(fast and slow)

1. Let fast pointer to move n steps in advance, making sure there is n steps gap between fast and slow

2. Move fast and slow pointer together until fast.next == null

code

 /**
  * Definition for singly-linked list.
  * public class ListNode {
  *     int val;
  *     ListNode next;
  *     ListNode(int x) { val = x; }
  * }
  */

 /*
 Time: O(n)
 Space: O(1)
 */
 class Solution {
       public ListNode removeNthFromEnd(ListNode head, int n) {
         ListNode dummy = new ListNode(-1);
         dummy.next = head;
         ListNode slow = dummy, fast = dummy;

         for (int i = 0; i < n; i++)  // fast先走n步
             fast = fast.next;

         while(fast.next != null) { // fast和slow一起走
             slow = slow.next;
             fast = fast.next;
         }
         //直接skip要删除的节点
         slow.next = slow.next.next;  // 思考为何不能写成 slow.next = fast;
         return dummy.next;
     }
 }