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) {
            return null;
        }

        // write code here
        ListNode p0 = head;
        ListNode p1 = null;
        ListNode p2 = null;
        ListNode head2 = null;

        while (p0 != null) {
            if (p1 == null) {
                p1 = p0;
            } else {
                p1.next = p0;
                p1 = p1.next;
            }
            p0 = p0.next;
            p1.next = null;

            if (p0 != null) {
                if (p2 == null) {
                    p2 = p0;
                    head2 = p2;
                } else {
                    p2.next = p0;
                    p2 = p2.next;
                }
                p0 = p0.next;
                p2.next = null;
            }
        }

        p1.next = head2;

        return head;
    }
}

五一来个链表排列，注意置空临时链表p1,p2尾指针，不然可能会stop the word & out memory error!#牛客AI配图神器#