将链表以m长度为一组反转链表

一群友分享的阿里面试题

 1.将head链表以m为组反转链表(不足m则不反转)：
 例子：假设m=3. 链表 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 ->8
 output: 3 -> 2 -> 1 ->6 ->5 ->4 -> 7 ->8

解法一：使用JDK的链表LinkedList

没有体现出链表的链式处理方式

代码

    public static void main(String[] args) {
        Integer[] array = new Integer[]{1, 2, 3, 4, 5, 6, 7, 8};
        LinkedList<Integer> linkedList = new LinkedList();
        linkedList.addAll(Arrays.asList(array));

        int m = 3;
        System.out.println("当 m = " + m);

        System.out.println("before handler");
        for (Integer integer : linkedList) {
            System.out.print(integer + " ");
        }
        System.out.println();

        //核心代码处
        for (int i = 0; i < linkedList.size(); i++) {
            if (i % m == m - 1) {
                int start = i - m + 1;
                int end = i;
                for (int a = 0; a < m / 2; a++) {
                    Integer temp = linkedList.get(start + a);
                    linkedList.set(start + a, linkedList.get(end - a));
                    linkedList.set(end - a, temp);
                }
            }
        }

        System.out.println("after handler");
        for (Integer integer : linkedList) {
            System.out.print(integer + " ");
        }
    }

运行效果：

当 m = 3
before handler
1 2 3 4 5 6 7 8
after handler
3 2 1 6 5 4 7 8

解法二：自己构造链表，用链的方式遍历处理链表

创建链表作为一个类使用

class Node<T> {
    T value;
    Node<T> next;
}

遍历链表

    public static <T> void printLink(Node<T> head) {
        Node<T> node = head;
        while (node != null) {
            System.out.print(node.value + "  ");
            node = node.next;
        }
        System.out.println();
    }

将数组转换成链表，并返回链表head

    /**
     * 将数组转成链表
     *
     * @param array
     * @param T类型的数组
     * @return Node<T>类型的链表head
     */
    public static <T> Node<T> arrayToNodeList(T[] array) {
        Node<T> node = new Node<T>();
        Node<T> head = node;
        for (T object : array) {
            Node<T> temp = new Node<T>();
            temp.value = object;
            node.next = temp;
            node = temp;
        }
        head = head.next;
        return head;
    }

递归分组反转链表

   /**
     * @param head 需要反转的头结点，
     * @param m    以m为一组反转
     * @return 返回以m为一组反转完成的链表
     */
    public static <T>Node<T> linkReverse(Node<T> head, int m) {
        if (head == null || head.next == null) {
            return head;
        }
        int count = 0;
        Node<T> current = head;
        while (count != m && current != null) {
            current = current.next;
            count++;
        }
        if (count == m) {
            current = linkReverse(current, m);
            while (count-- > 0) {
                Node<T> temp = head.next;
                head.next = current;
                current = head;
                head = temp;
            }
            head = current;
        }
        return head;
    }

实例测试

    public static void main(String[] args) {
        Integer[] array = new Integer[]{1, 2, 3, 4, 5, 6, 7, 8};
        //构造单向链表
        Node<Integer> head = arrayToNodeList(array);

        System.out.println("before handler");
        printLink(head);

        System.out.println("after handler");
        printLink(linkReverse(head, 3));
    }

结果输出

before handler
1  2  3  4  5  6  7  8
after handler
3  2  1  6  5  4  7  8