字节-LeetCode【148. 排序链表】

class Solution {

    /**
     * 二分法：时间复杂度 O(nlogn), 空间复杂度O(1)
     * @param head
     * @return
     */
    public ListNode sortList(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }

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

        // 截断分为两部分链表
        //  右部分节点头节点
        ListNode rightHead = slow.next;
        // 左部分头结点即为head
        slow.next = null;

        // 递归截断， 直至变为1个节点
        ListNode left = sortList(head);
        ListNode right = sortList(rightHead);

        //  归并排序
        ListNode node = new ListNode(-1);
        ListNode resultHead = node;
        while (left != null && right != null) {
            if (left.val  right.val) {
                node.next = left;
                left = left.next;
                node = node.next;
            } else {
                node.next = right;
                node = node.next;
                right = right.next;
            }
        }

        node.next = left == null ? right : left;

        return resultHead.next;

    }



    /**
     * 时间复杂度： On, 空间复杂度On
     * @param head
     * @return
     */
    public ListNode sortList2(ListNode head) {
        if (head == null) {
            return head;
        }

        List list = new ArrayList();
        while (head != null) {
            list.add(head.val);
            head = head.next;
        }

        Collections.sort(list, new Comparator() {
            @Override
            public int compare(Integer o1, Integer o2) {
                return o1 - o2;
            }
        });

        ListNode node = new ListNode(-1);
        ListNode pHead = node;
        for (int i = 0; i ) {
            node.next = new ListNode(list.get(i));
            node = node.next;
        }

        return pHead.next;

    }
}