题解 | #链表的奇偶重排# O(1)空间O(n)时间

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) {
            return head;
        }
        ListNode oodPointer = head;
        ListNode evenPointer = head.next;
        ListNode oodHead = oodPointer;
        ListNode evenHead = evenPointer;
        ListNode p1 = oodPointer;
        ListNode p2 = evenPointer;
        while (oodPointer != null || evenPointer != null) {
            if (oodPointer != null) {
                oodPointer = oodPointer.next;
                if (oodPointer != null) {
                    oodPointer = oodPointer.next;
                }
                p1.next = oodPointer;
                if (oodPointer != null) {
                    p1 = p1.next;
                }
            }
            if (evenPointer != null) {
                evenPointer = evenPointer.next;
                if (evenPointer != null) {
                    evenPointer = evenPointer.next;
                }
                p2.next = evenPointer;
                if (evenPointer != null) {
                    p2 = p2.next;
                }
            }
        }
        if (p2 != null) {
            p2.next = null;
        }
        p1.next = evenHead;
        return oodHead;
    }
}