Leetcode 148. Sort List

Sort a linked list in O(n log n) time using constant space complexity.

题意：常数级别的空间复杂度、O(n log n) 的时间复杂度对链表排序。
思路：对链表进行归并排序。

1. 待排序的链表如果是null或者只有一个节点，则直接返回。
2. 找到链表的中间节点。
3. 对左右两段继续递归进行排序。
4. 将排序好的两段合并成一个链表。

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

    //找middle
    ListNode middle = findMiddle(head);

    //拆两段
    ListNode right = middle.next;
    middle.next = null;

    //对两段分别再进行sort
    ListNode left = sortList(head);
    right = sortList(right);

    //merge两段
    return merge(left, right);
}

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

    return slow;
}

public ListNode merge(ListNode left, ListNode right) {
    ListNode root = new ListNode(0);
    ListNode dummy = root;
    while (left != null || right != null) {
        if (left == null) {
            dummy.next = right;
            right = right.next;
        } else if (right == null) {
            dummy.next = left;
            left = left.next;
        } else {
            if (left.val < right.val) {
                dummy.next = left;
                left = left.next;
            } else {
                dummy.next = right;
                right = right.next;
            }
        }
        dummy = dummy.next;
    }
    return dummy.next;
}