题解 | #链表的奇偶重排#
链表的奇偶重排
https://www.nowcoder.com/practice/02bf49ea45cd486daa031614f9bd6fc3
1、遍历链表,给一个指针记录节点位数即可
2、采用链表的变量副本来表示链表上的当前指针
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 ListNode oddNumber = new ListNode(-1); ListNode oddCurr = oddNumber; ListNode evenNumber = new ListNode(-1); ListNode evenCurr = evenNumber; int p = 1; while (head != null) { int number = head.val; if (p % 2 == 0) { // 偶数 evenCurr.next = new ListNode(number); evenCurr = evenCurr.next; } else { // 奇数 oddCurr.next = new ListNode(number); oddCurr = oddCurr.next; } head = head.next; p++; } oddCurr.next = evenNumber.next; return oddNumber.next; } }