题解 | #链表的奇偶重排#

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;
    }
}