题解 | #牛的品种排序IV#
牛的品种排序IV
https://www.nowcoder.com/practice/bd828af269cd493c86cc915389b02b9f
知识点:链表 遍历
模拟思路:没理解这题的想用的算法,单纯的遍历链表,筛选出0,1,放一起感觉没啥意义,但是复杂度O(n),已经最快了;如果这就是答案,那这算不上困难题,感觉牛客题库不太行(不是)
编程语言:java
如果我的思路启发了你,给个小小关注吧~
我是废江,一个从java跑到内核再准备润回java的打工人,我会持续分享从linux内核到上层java微服务等干货
import java.util.*; /* * public class ListNode { * int val; * ListNode next = null; * public ListNode(int val) { * this.val = val; * } * } */ public class Solution { /** * 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可 * * * @param head ListNode类 * @return ListNode类 */ public ListNode sortCowsIV (ListNode head) { // write code here if(head.next == null) return head; ListNode l1 = new ListNode(-1); ListNode cur = l1; ListNode l2 = new ListNode(-1); ListNode cur2 =l2; ListNode tmp = null; ListNode flag = null; while (head != null) { if (head.val == 0) { tmp = new ListNode(0); tmp.next = null; l1.next = tmp; l1 = tmp; flag = l1; } else { tmp = new ListNode(1); tmp.next = null; l2.next = tmp; l2 = tmp; } head = head.next; } flag.next = cur2.next; return cur.next; } }