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) { // write code here // 定一个两个链表, 分别存放奇数位链表值和偶数位链表值 ListNode odd = new ListNode(-1); ListNode tailOdd = odd; ListNode even = new ListNode(-1); ListNode tailEven = even; // 第一个位置则为奇数位 int index = 1; while (head != null) { if (index % 2 == 1) { tailOdd.next = head; tailOdd = tailOdd.next; } else { tailEven.next = head; tailEven = tailEven.next; } head = head.next; index ++; } tailOdd.next = even.next; tailEven.next = null; return odd.next; } }