题解 | #链表的奇偶重排#

分拆链表

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) {
        ListNode p1 = new ListNode(-1);
        ListNode p2 = new ListNode(-1);
        ListNode p11 = p1;
        ListNode p22 = p2;

        if (head == null) {
            return head;
        }

        int i = 0;
        while( head != null) {
            ListNode next = head.next;
            head.next = null;
            if (i % 2 == 1) { // 偶数
                p22.next = head;
                p22 = p22.next;
            } else {
                p11.next = head;
                p11 = p11.next;
            }
            head = next;
            i++;
        }
        p11.next = p2.next;
        return p1.next;
    }
}