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) {
        if (head == null || head.next == null) {
            return head;
        }
        ListNode current1 = head;
        ListNode current2 = head.next;
        ListNode start2 = head.next;
        ListNode preNode1 = null;
        ListNode preNode2 = null;
        while (current1 != null || current2 != null) {
            if (preNode1 != null) {
                preNode1.next = current1;
            }
            if (preNode2 != null) {
                preNode2.next = current2;
            }
            preNode1 = current1;
            preNode2 = current2;
            if (current1 != null) {
                current1 = current1.next == null ? null : current1.next.next;
            }
            if (current2 != null) {
                current2 = current2.next == null ? null : current2.next.next;
            }
        }
        preNode1.next = start2;
        return head;
    }
}