Remove Nth Node From End of List

最新推荐文章于 2024-06-20 12:29:29 发布

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最新推荐文章于 2024-06-20 12:29:29 发布

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原文链接：http://www.cnblogs.com/qq1029579233/p/4478009.html

本文介绍了一个算法，用于从链表中删除倒数第N个节点，并通过实例演示实现过程。

https://leetcode.com/problems/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.

 1 public class Solution {
 2     public static ListNode removeNthFromEnd(ListNode head, int n) {
 3     ListNode li=head;
 4         int len=0;
 5         while(li!=null){
 6             len++;
 7             li=li.next;
 8         }
 9         int index=len-n+1;
10         if(index==1){return head.next;}
11         li=head;
12         ListNode pre_li=li;
13         while(index-->1){
14             pre_li=li;
15             li=li.next;
16         }
17         pre_li.next=li.next;
18         return head;
19     }
20 //    public static class ListNode {
21 //    int val;
22 //    ListNode next;
23 //
24 //    ListNode(int x) {
25 //        val = x;
26 //    }
27 //    }
28     public static void main(String[]args){
29     ListNode[]node=new ListNode[4];
30     for(int i=0;i<node.length;i++){
31         node[i]=new ListNode(i);
32     }
33     node[0].next=node[1];
34     node[1].next=node[2];
35     node[2].next=node[3];
36     ListNode head=removeNthFromEnd(node[0],2);
37     while(head!=null){
38         System.out.println(head.val);
39         head=head.next;
40     }
41     }
42 }