148. Sort List | Java最短代码实现

原题链接： 148. Sort List

【思路】

本题考查归并排序。由于没有random access指针，那么显然是不是合快排。归并排序的基本思想：找到链表的mid节点，然后递归前半部分和后半部分分别进行归并排序，每次排完序之后都要对排好序的链表进行归并：

    public ListNode sortList(ListNode head) {
        if (head  == null || head.next == null) return head;
        ListNode mid = getMidListNode(head);
        ListNode list2 = mid.next;
        mid.next = null;
        return mergeList(sortList(head), sortList(list2));
    }
    private ListNode mergeList(ListNode list1, ListNode list2) {
        ListNode dummyHead = new ListNode(0);
        ListNode curr = dummyHead;
        while (list1 != null && list2 != null) {
            if (list1.val < list2.val) {
                curr.next = list1;
                list1 = list1.next;
            } else {
                curr.next = list2;
                list2 = list2.next;
            }
            curr = curr.next;
        }
        curr.next = list1 != null ? list1 : list2;
        return dummyHead.next;
    }
    private ListNode getMidListNode(ListNode head) {
        ListNode slow = head, fast = head;
        while (fast.next != null && fast.next.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }

15 / 15  test cases passed. Runtime: 8 ms  Your runtime beats 42.29% of javasubmissions.

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