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
        ListNode vHead = new ListNode(0);
        ListNode cur, newList;
        newList = vHead;

        if(head==null||head.next==null){
            return head;
        }
        
        cur = head;  // odd
        while(true){
            System.out.println(cur.val);
            ListNode tail = new ListNode(cur.val);
            vHead.next = tail;
            vHead = tail;
            if(cur.next==null||cur.next.next==null){  // 在调用指针的时候，如果已经为null，那么再.next会抛出越界报错
                break;
            }
            cur = cur.next.next;
        }

        cur = head.next;
        while(true){
            System.out.println(cur.val);
            ListNode tail = new ListNode(cur.val);
            vHead.next = tail;
            vHead = tail;
            if(cur.next==null||cur.next.next==null){
                break;
            }
            cur = cur.next.next;
        }
        

        return newList.next;
    }
}

在调用指针的时候，如果已经为null，那么再.next会抛出越界报错