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 || head.next.next == null){
            return head;
        }

        ListNode oHead = head.next, jCur = head, oCur = oHead;
        while(oCur.next != null && oCur.next.next != null){
            jCur.next = jCur.next.next;
            jCur = jCur.next;
            oCur.next = oCur.next.next;
            oCur = oCur.next;
            System.out.println(jCur.val);
        }

        //链表数量为奇数的时候，最后还有个奇数位节点
        if(oCur.next != null){
            jCur.next = oCur.next;
            //大坑啊，偶数位最后一个要与最后一个奇数位断开，一定要断开，不然就成环了。
            oCur.next = null;
            jCur = jCur.next;
        }

        jCur.next = oHead;

        return head;
    }
}