题解 | #链表的奇偶重排#

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;
    }
}