链表反序(C#版)

有次看见别人的面试题问起链表反序.自己实验下.

 1    public class Node<T> : ICloneable
 2    {
 3        public T Data { get; set; }
 4
 5        public Node<T> Next { get; set; }
 6        /**//// <summary>
 7        /// 在遍历当前链表时,把遍历的当前节点赋给新的节点,新的节点会记录当前数据,加进新链表的前头,算法为O(N)次.
 8        /// </summary>
 9        /// <param name="node"></param>
10        /// <returns></returns>
11        public static Node<T> reverse(Node<T> node)
12        {
13            Node<T> head = null;
14            while (node != null)
15            {
16                Node<T> newNode = new Node<T>();
17                //设置新节点的数据为当前的节点的数据,并连接head
18                newNode.Data = node.Data;
19                newNode.Next = head;
20                //head记录头节点
21                head = newNode;
22
23                //遍历下一个节点
24                node = node.Next;                            
25            }
26            return head;
27        }
28        public object Clone()
29        {
30            Node<T> newnode = new Node<T>();
31            newnode.Data = this.Data;
32            return newnode;
33        }
34    }

 在主程序下:

 1        static void Test1()
 2        {           
 3            Node<string> s6 = new Node<string> { Data="6"};
 4            Node<string> s5 = new Node<string> { Data = "5", Next = s6 };
 5            Node<string> s4 = new Node<string> { Data = "4", Next = s5 };
 6            Node<string> s3 = new Node<string> { Data = "3", Next = s4 };
 7            Node<string> s2 = new Node<string> { Data = "2", Next = s3 };
 8            Node<string> s1 = new Node<string> { Data = "1", Next = s2 };
 9
10            Node<string> head = Node<string>.reverse(s1);
11            
12
13            while (head != null)
14            {
15                Console.Write(head.Data);
16                head = head.Next;
17            }
18            Console.Read();
19        }

十进制与别的进制相互转化：

       /// <summary>
        /// 把相关十进制整数转成x进制
        /// </summary>
        /// <param name="i">十进制整数</param>
        /// <param name="x">x进制</param>
        /// <returns></returns>
        public int DOI(int i, int x)
        {         
            int result = 0;
            int j = 0;
            while (i > 0)
            {                
                result += (int)Math.Pow(10,j)*(i % x);
                i = i / x;
                j++;
            }
            return result;
        }
        /// <summary>
        /// 把相关x进制整数转成十进制
        /// </summary>
        /// <param name="i">x进制整数</param>
        /// <param name="x">x进制</param>
        /// <returns></returns>
        public int IOD(int i, int x)
        {
            int result = 0;
            int j = 0;
            while (i > 0)
            {
                result += (int)Math.Pow(x, j) * (i % 10);
                i = i / 10;
                j++;
            }
            return result;
        }
        /// <summary>
        /// 根据上面二段推出x进制的数转成y进制
        /// </summary>
        /// <param name="i"></param>
        /// <param name="x"></param>
        /// <param name="y"></param>
        /// <returns></returns>
        public int IOI(int i, int x, int y)
        {
            int result = 0;
            int j = 0;
            while (i > 0)
            {
                result += (int)Math.Pow(x, j) * (i % y);
                i = i / y;
                j++;
            }
            return result;

        }