C#，电话数字键盘问题（Mobile Numeric Keypad problem）的算法与源代码

1 电话数字键盘问题

提供移动数字键盘。您只能按向上、向左、向右或向下至当前按钮的按钮。不允许您按最下面一行的角点按钮（即.*和#）。

移动键盘

给定一个数N，找出给定长度的可能数。

示例：

对于N=1，可能的数字数为10（0、1、2、3、…、9）

对于N=2，可能的数字数为36

可能的数字：00、08、11、12、14、22、21、23、25等等。

如果我们从0开始，有效数字将是00、08（计数：2）

如果我们从1开始，有效数字将是11、12、14（计数：3）

如果我们从2开始，有效数字将是22、21、23、25（计数：4）

如果我们从3开始，有效数字将是33、32、36（计数：3）

如果我们从4开始，有效数字将是44,41,45,47（计数：4）

如果我们从5开始，有效数字将是55,54,52,56,58（计数：5）

………………………………

我们需要打印可能的数字。

推荐做法

移动数字键盘

试试看！

N=1是普通情况，可能的数字数为10（0，1，2，3，…，9）

对于N>1，我们需要从某个按钮开始，然后移动到四个方向（向上、向左、向右或向下）中的任何一个，该方向指向有效的按钮（不应转到*、#）。继续这样做，直到获得N个长度编号（深度优先遍历）。

递归解决方案：

移动键盘是4X3的矩形网格（4行3列）

假设Count（i，j，N）表示从位置（i，j）开始的N个长度数字的计数

如果N=1 ,计数（i，j，N）=10

其他的 Count（i，j，N）=所有Count（r，c，N-1）之和，其中（r，c）是新的

长度1从当前有效移动后的位置

位置（i，j）

下面是上述公式的三种实现代码。

2 源代码

using System;
using System.Collections;
using System.Collections.Generic;namespace Legalsoft.Truffer.Algorithm
{public static partial class Algorithm_Gallery{#region 算法1private static int MNKP_Solve_Utility(char[,] keypad, int i, int j, int n){if (keypad == null || n <= 0){return 0;}if (n == 1){return 1;}int[] row = { 0, 0, -1, 0, 1 };int[] col  = { 0, -1, 0, 1, 0 };int totalCount = 0;for (int move = 0; move < 5; move++){int ro = i + row[move];int co = j + col[move];if (ro >= 0 && ro <= 3 && co >= 0 && co <= 2 && keypad[ro, co] != '*' && keypad[ro, co] != '#'){totalCount += MNKP_Solve_Utility(keypad, ro, co, n - 1);}}return totalCount;}public static int MNKP_Solve(char[,] keypad, int n){if (keypad == null || n <= 0){return 0;}if (n == 1){return 10;}int totalCount = 0;for (int i = 0; i < 4; i++) {for (int j = 0; j < 3; j++) {if (keypad[i, j] != '*' && keypad[i, j] != '#'){totalCount += MNKP_Solve_Utility(keypad, i, j, n);}}}return totalCount;}#endregion#region 算法2public static int MNKP_Solve_2th(char[,] keypad, int n){if (keypad == null || n <= 0){return 0;}if (n == 1){return 10;}int[] row = { 0, 0, -1, 0, 1 };int[] col = { 0, -1, 0, 1, 0 };int[,] count = new int[10, n + 1];for (int i = 0; i < 10; i++){count[i, 0] = 0;count[i, 1] = 1;}for (int k = 2; k <= n; k++){for (int i = 0; i < 4; i++) {for (int j = 0; j < 3; j++) {if (keypad[i, j] != '*' && keypad[i, j] != '#'){int num = keypad[i, j] - '0';count[num, k] = 0;for (int move = 0; move < 5; move++){int ro = i + row[move];int co = j + col[move];if (ro >= 0 && ro <= 3 && co >= 0 && co <= 2 && keypad[ro, co] != '*' && keypad[ro, co] != '#'){int nextNum = keypad[ro, co] - '0';count[num, k] += count[nextNum, k - 1];}}}}}}int totalCount = 0;for (int i = 0; i < 10; i++){totalCount += count[i, n];}return totalCount;}#endregion#region 算法3public static int MNTP_Solve_3th(char[,] keypad, int n){if (keypad == null || n <= 0){return 0;}if (n == 1){return 10;}int[] odd = new int[10];int[] even = new int[10];int useOdd = 0;for (int i = 0; i < 10; i++){odd[i] = 1;}for (int j = 2; j <= n; j++){useOdd = 1 - useOdd;if (useOdd == 1){even[0] = odd[0] + odd[8];even[1] = odd[1] + odd[2] + odd[4];even[2] = odd[2] + odd[1] + odd[3] + odd[5];even[3] = odd[3] + odd[2] + odd[6];even[4] = odd[4] + odd[1] + odd[5] + odd[7];even[5] = odd[5] + odd[2] + odd[4] + odd[8] + odd[6];even[6] = odd[6] + odd[3] + odd[5] + odd[9];even[7] = odd[7] + odd[4] + odd[8];even[8] = odd[8] + odd[0] + odd[5] + odd[7] + odd[9];even[9] = odd[9] + odd[6] + odd[8];}else{odd[0] = even[0] + even[8];odd[1] = even[1] + even[2] + even[4];odd[2] = even[2] + even[1] + even[3] + even[5];odd[3] = even[3] + even[2] + even[6];odd[4] = even[4] + even[1] + even[5] + even[7];odd[5] = even[5] + even[2] + even[4] + even[8] + even[6];odd[6] = even[6] + even[3] + even[5] + even[9];odd[7] = even[7] + even[4] + even[8];odd[8] = even[8] + even[0] + even[5] + even[7] + even[9];odd[9] = even[9] + even[6] + even[8];}}int totalCount = 0;for (int i = 0; i < 10; i++){totalCount += (useOdd == 1) ? even[i] : odd[i];}return totalCount;}#endregion}
}

3 源程序

using System;
using System.Collections;
using System.Collections.Generic;

namespace Legalsoft.Truffer.Algorithm
{
   public static partial class Algorithm_Gallery
   {
#region 算法1
       private static int MNKP_Solve_Utility(char[,] keypad, int i, int j, int n)
       {
           if (keypad == null || n <= 0)
           {
               return 0;
           }
           if (n == 1)
           {
               return 1;
           }

int[] row = { 0, 0, -1, 0, 1 };
int[] col = { 0, -1, 0, 1, 0 };

           int totalCount = 0;
           for (int move = 0; move < 5; move++)
           {
               int ro = i + row[move];
               int co = j + col[move];
               if (ro >= 0 && ro <= 3 && co >= 0 && co <= 2 && keypad[ro, co] != '*' && keypad[ro, co] != '#')
               {
                   totalCount += MNKP_Solve_Utility(keypad, ro, co, n - 1);
               }
           }
           return totalCount;
       }

       public static int MNKP_Solve(char[,] keypad, int n)
       {
           if (keypad == null || n <= 0)
           {
               return 0;
           }
           if (n == 1)
           {
               return 10;
           }

           int totalCount = 0;
           for (int i = 0; i < 4; i++)
           {
               for (int j = 0; j < 3; j++)
               {
                   if (keypad[i, j] != '*' && keypad[i, j] != '#')
                   {
                       totalCount += MNKP_Solve_Utility(keypad, i, j, n);
                   }
               }
           }
           return totalCount;
       }
       #endregion

       #region 算法2
       public static int MNKP_Solve_2th(char[,] keypad, int n)
       {
           if (keypad == null || n <= 0)
           {
               return 0;
           }
           if (n == 1)
           {
               return 10;
           }
           int[] row = { 0, 0, -1, 0, 1 };
           int[] col = { 0, -1, 0, 1, 0 };

int[,] count = new int[10, n + 1];

           for (int i = 0; i < 10; i++)
           {
               count[i, 0] = 0;
               count[i, 1] = 1;
           }

           for (int k = 2; k <= n; k++)
           {
               for (int i = 0; i < 4; i++)
               {
                   for (int j = 0; j < 3; j++)
                   {
                       if (keypad[i, j] != '*' && keypad[i, j] != '#')
                       {
                           int num = keypad[i, j] - '0';
                           count[num, k] = 0;

                           for (int move = 0; move < 5; move++)
                           {
                               int ro = i + row[move];
                               int co = j + col[move];
                               if (ro >= 0 && ro <= 3 && co >= 0 && co <= 2 && keypad[ro, co] != '*' && keypad[ro, co] != '#')
                               {
                                   int nextNum = keypad[ro, co] - '0';
                                   count[num, k] += count[nextNum, k - 1];
                               }
                           }
                       }
                   }
               }
           }

           int totalCount = 0;
           for (int i = 0; i < 10; i++)
           {
               totalCount += count[i, n];
           }
           return totalCount;
       }
       #endregion

       #region 算法3
       public static int MNTP_Solve_3th(char[,] keypad, int n)
       {
           if (keypad == null || n <= 0)
           {
               return 0;
           }
           if (n == 1)
           {
               return 10;
           }
           int[] odd = new int[10];
           int[] even = new int[10];

           int useOdd = 0;
           for (int i = 0; i < 10; i++)
           {
               odd[i] = 1;
           }

           for (int j = 2; j <= n; j++)
           {
               useOdd = 1 - useOdd;

               if (useOdd == 1)
               {
                   even[0] = odd[0] + odd[8];
                   even[1] = odd[1] + odd[2] + odd[4];
                   even[2] = odd[2] + odd[1] + odd[3] + odd[5];
                   even[3] = odd[3] + odd[2] + odd[6];
                   even[4] = odd[4] + odd[1] + odd[5] + odd[7];
                   even[5] = odd[5] + odd[2] + odd[4] + odd[8] + odd[6];
                   even[6] = odd[6] + odd[3] + odd[5] + odd[9];
                   even[7] = odd[7] + odd[4] + odd[8];
                   even[8] = odd[8] + odd[0] + odd[5] + odd[7] + odd[9];
                   even[9] = odd[9] + odd[6] + odd[8];
               }
               else
               {
                   odd[0] = even[0] + even[8];
                   odd[1] = even[1] + even[2] + even[4];
                   odd[2] = even[2] + even[1] + even[3] + even[5];
                   odd[3] = even[3] + even[2] + even[6];
                   odd[4] = even[4] + even[1] + even[5] + even[7];
                   odd[5] = even[5] + even[2] + even[4] + even[8] + even[6];
                   odd[6] = even[6] + even[3] + even[5] + even[9];
                   odd[7] = even[7] + even[4] + even[8];
                   odd[8] = even[8] + even[0] + even[5] + even[7] + even[9];
                   odd[9] = even[9] + even[6] + even[8];
               }
           }

           int totalCount = 0;
           for (int i = 0; i < 10; i++)
           {
               totalCount += (useOdd == 1) ? even[i] : odd[i];
           }
           return totalCount;
       }
#endregion
}
}