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 odd = head;
        ListNode even = head.next;
        ListNode oddHead = odd , evenHead = even;
        while(even != null && even.next != null){
            odd.next = even.next;    //此处写成odd.next = odd.next.next；时间复杂度会超，离谱；
            odd = odd.next;
            even.next = odd.next;    //even.next = even.next.next；会超；
            even = even.next;
        }
        odd.next = evenHead;
        return oddHead;
    }
}