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
        if(head==null||head.next==null){
            return head;
        }
        ListNode res=new ListNode(0);
        ListNode p=res;
        ListNode p1=head;
        ListNode p2=head;
        int count=1;
        while(p1!=null){
            if((count&1)==1){
                p.next=new ListNode(p1.val);
                p=p.next;
            }
            p1=p1.next;
            count++;
        }
        count=1;
        while(p2!=null){
            if((count&1)==0){
                p.next=new ListNode(p2.val);
                p=p.next;
            }
            p2=p2.next;
            count++;
        }
        return res.next;
    }
}