【leetcode刷题笔记】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.

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题解：快慢指针的思想，就像找到链表中点或者判断链表是否有环一样，是很经典的思想。

设置fast指针比slow指针多走n步，然后slow和fast一起前进，当fast指向null的时候，slow就指向从后往前数的第n个元素了。

但是要删除某个节点，最好的是知道它前面的节点，所以slow实际比fast慢n+1步。

代码如下：

 /**
  * 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)
             return null;

         ListNode slow = new ListNode(0);
         ListNode answer = slow;
         slow.next = head;
         ListNode fast = head;
         for(int i = 0;i < n;i++){
             if(fast == null)
                 return null;
             fast = fast.next;
         }
         while(fast != null){
             slow = slow.next;
             fast = fast.next;
         }

         //delete what slow.next
         slow.next = slow.next.next;
         return answer.next;
     }
 }