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 n1 = new ListNode(-1);
        ListNode n2 = new ListNode(-1);
        ListNode res = n1;
        ListNode re2 = n2;
        int index = 1;
        while (head != null) {
            if (index % 2 == 1) {
                n1.next = head;
                n1 = n1.next;
            } else {
                n2.next = head;
                n2 = n2.next;
            }
            index++;
            head = head.next;
        }
         n2.next = null;
       // if (n1 != null) {
             n1.next = re2.next;
      //  }
       
       
        return res.next;
    }
}

遍历将奇偶数据拆分位两个链表，再次组合