题解 | #链表的奇偶重排#
链表的奇偶重排
https://www.nowcoder.com/practice/02bf49ea45cd486daa031614f9bd6fc3
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 oddEvenList (ListNode head) { if(head == null || head.next == null || head.next.next == null){ return head; } ListNode oHead = head.next, jCur = head, oCur = oHead; while(oCur.next != null && oCur.next.next != null){ jCur.next = jCur.next.next; jCur = jCur.next; oCur.next = oCur.next.next; oCur = oCur.next; System.out.println(jCur.val); } //链表数量为奇数的时候,最后还有个奇数位节点 if(oCur.next != null){ jCur.next = oCur.next; //大坑啊,偶数位最后一个要与最后一个奇数位断开,一定要断开,不然就成环了。 oCur.next = null; jCur = jCur.next; } jCur.next = oHead; return head; } }#学习##刷题##每日刷题#