奇偶升降序链表排序

看到一道题目，还挺有意思

一个链表奇数位上升序，偶数位上降序，不用额外空间让这个链表整体升序，例如：1 8 3 6 5 4 7 2 9，最后输出1 2 3 4 5 6 7 8 9。

其实不算很难，就是题目比较新颖

解：首先分离出奇数链表和偶数链表，偶数链表反转一下，就成为升序了，最后把两条升序链表合并即可（leetcode上这种题也挺经典的）

public class SortList {

    class ListNode {
        int val;
        ListNode next;

        public ListNode(int val) {
            this.val = val;
        }

        public ListNode(int val, ListNode next) {
            this.val = val;
            this.next = next;
        }

        @Override public String toString() {
            return "ListNode{" + "val=" + val + ", next=" + next + '}';
        }
    }

    @Test
    public void test() {
        //测试 1-8-3-6-5-4-7-2-9
        ListNode l9 = new ListNode(9);
        ListNode l2 = new ListNode(2, l9);
        ListNode l7 = new ListNode(7, l2);
        ListNode l4 = new ListNode(4, l7);
        ListNode l5 = new ListNode(5, l4);
        ListNode l6 = new ListNode(6, l5);
        ListNode l3 = new ListNode(3, l6);
        ListNode l8 = new ListNode(8, l3);
        ListNode head = new ListNode(1, l8);
        ListNode ans = sort(head);
        System.out.println(ans);
    }

    public ListNode sort(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        ListNode head1 = head, head2 = head.next, pre = head1, cur = head2;
        //将奇数偶数链表拆分出来
        while (cur != null && cur.next != null) {
            ListNode next = cur.next;
            cur.next = next.next;
            pre.next = next;
            pre = pre.next;
            cur = cur.next;
        }
        head2 = reverse(head2);
        return merge(head1, head2).next;
    }

    //反转链表
    private ListNode reverse(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        ListNode root = reverse(head.next);
        head.next.next = head;
        head.next = null;
        return root;
    }

   //合并两条链表，升序
    private ListNode merge(ListNode head1, ListNode head2) {
        ListNode dummy = new ListNode(0), tmp = dummy;
        while (head1 != null && head2 != null) {
            if (head1.val < head2.val) {
                tmp.next = head1;
                head1 = head1.next;
            } else {
                tmp.next = head2;
                head2 = head2.next;
            }
            tmp = tmp.next;
        }
        if (head1 != null) {
            tmp.next = head1;
        }
        if (head2 != null) {
            tmp.next = head2;
        }
        return dummy;
    }
}

标签：奇偶,head,ListNode,head2,val,next,链表,head1,排序
来源： https://www.cnblogs.com/zjytrhy/p/15055211.html