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) {
        if (head == null || head.next == null) {
            return head;
        }
        ListNode head2 = head.next;
        // write code here
        ListNode start = head;
        ListNode start2 = head;
        ListNode befstart2 = start2;
        while (true) {
            start = start2.next;
            if (start == null) {
                break;
            }
            start2.next = start.next;
            befstart2 = start2;
            start2 = start.next;
            if (start2 == null) {
                break;
            }
            start.next = start2.next;
        }
        if (start2 == null) {
            befstart2.next = head2;
        }
        if (start == null) {
            start2.next = head2;
        }
        return head;
    }
}

空间复杂度不需要O（n）,也不用判断链表是奇数还是偶数就能求解