题目

输入一个链表的头节点，请将该链表排序。

分析

归并排序的主要思想是将链表分成两个子链表，在对两个子链表排序后再将它们合并成一个排序的链表。
这里可以用快慢双指针的思路将链表分成两半。如果慢指针一次走一步，快指针一次走两步，当快指针走到链表尾部时，慢指针只走到链表的中央，这样也就找到了链表后半部分的头节点。

解

public class Test {
    public static void main(String[] args) {
        ListNode listNode1 = new ListNode(1);
        ListNode listNode2 = new ListNode(2);
        ListNode listNode3 = new ListNode(3);
        ListNode listNode4 = new ListNode(4);
        ListNode listNode5 = new ListNode(5);
        ListNode listNode6 = new ListNode(6);

        listNode3.next = listNode5;
        listNode5.next = listNode1;
        listNode1.next = listNode4;
        listNode4.next = listNode2;
        listNode2.next = listNode6;

        ListNode result = sortList(listNode3);
        while (result != null) {
            System.out.println(result.val);
            result = result.next;
        }
    }

    public static ListNode sortList(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }

        ListNode head1 = head;
        ListNode head2 = split(head);

        head1 = sortList(head1);
        head2 = sortList(head2);

        return merge(head1, head2);
    }

    private static ListNode split(ListNode head) {
        ListNode slow = head;
        ListNode fast = head.next;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }

        ListNode second = slow.next;
        slow.next = null;

        return second;
    }

    private static ListNode merge(ListNode head1, ListNode head2) {
        ListNode dummy = new ListNode(0);
        ListNode cur = dummy;
        while (head1 != null && head2 != null) {
            if (head1.val < head2.val) {
                cur.next = head1;
                head1 = head1.next;
            }
            else {
                cur.next = head2;
                head2 = head2.next;
            }

            cur = cur.next;
        }
        cur.next = head1 == null ? head2 : head1;
        return dummy.next;
    }
}