【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：

 1 /**

 2  * Definition for singly-linked list.

 3  * public class ListNode {

 4  *     int val;

 5  *     ListNode next;

 6  *     ListNode(int x) {

 7  *         val = x;

 8  *         next = null;

 9  *     }

10  * }

11  */

12 public class Solution {

13     public ListNode removeNthFromEnd(ListNode head, int n) {

14         if(head == null)

15             return null;

16         

17         ListNode slow = new ListNode(0);

18         ListNode answer = slow;

19         slow.next = head;

20         ListNode fast = head;

21         for(int i = 0;i < n;i++){

22             if(fast == null)

23                 return null;

24             fast = fast.next;

25         }

26         while(fast != null){

27             slow = slow.next;

28             fast = fast.next;

29         }

30         

31         //delete what slow.next

32         slow.next = slow.next.next;

33         return answer.next;

34     }

35 }