手机数字键盘问题详细介绍

2021年5月4日22:55:42 发表评论 895 次浏览

本文概述

给定移动数字小键盘。你只能按向上, 向左, 向右或向下按当前按钮。你不允许按底行的角按钮(即*和＃)。

手机键盘

给定数字N, 找出给定长度的可能数字。

例子：

对于N = 1, 可能的数量为10(0、1、2、3, ...., 9)

对于N = 2, 可能的数量为36

可能的数字：00, 08 11, 12, 14 22, 21, 23, 25, 依此类推。

如果我们以0开头, 则有效数字将为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)

如果我们以3开头, 则有效数字将为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。

N = 1是微不足道的情况, 可能的数量为10(0, 1, 2, 3, …。, 9)

对于N> 1, 我们需要从某个按钮开始, 然后移至四个方向(向上, 向左, 向右或向下)中的任意一个, 然后转到一个有效按钮(不应转到*, ＃)。继续执行此操作, 直到获得N个长度数字(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)代表从位置(i, j)开始的N个长度数字的计数

If N = 1
  Count(i, j, N) = 10  
Else
  Count(i, j, N) = Sum of all Count(r, c, N-1) where (r, c) is new 
                   position after valid move of length 1 from current 
                   position (i, j)

以下是上述递归公式的实现。

C ++

//A Naive Recursive C program to count number of possible numbers
//of given length
#include <stdio.h>
  
//left, up, right, down move from current location
int row[] = {0, 0, -1, 0, 1};
int col[] = {0, -1, 0, 1, 0};
  
//Returns count of numbers of length n starting from key position
//(i, j) in a numeric keyboard.
int getCountUtil( char keypad[][3], int i, int j, int n)
{
     if (keypad == NULL || n <= 0)
         return 0;
  
     //From a given key, only one number is possible of length 1
     if (n == 1)
         return 1;
  
     int k=0, move=0, ro=0, co=0, totalCount = 0;
  
     //move left, up, right, down from current location and if
     //new location is valid, then get number count of length
     //(n-1) from that new position and add in count obtained so far
     for (move=0; move<5; move++)
     {
         ro = i + row[move];
         co = j + col[move];
         if (ro>= 0 && ro <= 3 && co>=0 && co <= 2 &&
            keypad[ro][co] != '*' && keypad[ro][co] != '#' )
         {
             totalCount += getCountUtil(keypad, ro, co, n-1);
         }
     }
  
     return totalCount;
}
  
//Return count of all possible numbers of length n
//in a given numeric keyboard
int getCount( char keypad[][3], int n)
{
     //Base cases
     if (keypad == NULL || n <= 0)
         return 0;
     if (n == 1)
         return 10;
  
     int i=0, j=0, totalCount = 0;
     for (i=0; i<4; i++)  //Loop on keypad row
     {
         for (j=0; j<3; j++)   //Loop on keypad column
         {
             //Process for 0 to 9 digits
             if (keypad[i][j] != '*' && keypad[i][j] != '#' )
             {
                 //Get count when number is starting from key
                 //position (i, j) and add in count obtained so far
                 totalCount += getCountUtil(keypad, i, j, n);
             }
         }
     }
     return totalCount;
}
  
//Driver program to test above function
int main( int argc, char *argv[])
{
    char keypad[4][3] = {{ '1' , '2' , '3' }, { '4' , '5' , '6' }, { '7' , '8' , '9' }, { '*' , '0' , '#' }};
    printf ( "Count for numbers of length %d: %dn" , 1, getCount(keypad, 1));
    printf ( "Count for numbers of length %d: %dn" , 2, getCount(keypad, 2));
    printf ( "Count for numbers of length %d: %dn" , 3, getCount(keypad, 3));
    printf ( "Count for numbers of length %d: %dn" , 4, getCount(keypad, 4));
    printf ( "Count for numbers of length %d: %dn" , 5, getCount(keypad, 5));
  
    return 0;
}

Java

//A Naive Recursive Java program 
//to count number of possible 
//numbers of given length
class GfG
{
  
//left, up, right, down 
//move from current location
static int row[] = { 0 , 0 , - 1 , 0 , 1 };
static int col[] = { 0 , - 1 , 0 , 1 , 0 };
  
//Returns count of numbers of length
//n starting from key position
//(i, j) in a numeric keyboard.
static int getCountUtil( char keypad[][], int i, int j, int n)
{
     if (keypad == null || n <= 0 )
         return 0 ;
  
     //From a given key, only one 
     //number is possible of length 1
     if (n == 1 )
         return 1 ;
  
     int k = 0 , move = 0 , ro = 0 , co = 0 , totalCount = 0 ;
  
     //move left, up, right, down
     //from current location and if
     //new location is valid, then 
     //get number count of length
     //(n-1) from that new position 
     //and add in count obtained so far
     for (move= 0 ; move<5 ; move++)
     {
         ro = i + row[move];
         co = j + col[move];
         if (ro>= 0 && ro <= 3 && co>= 0 && co <= 2 &&
         keypad[ro][co] != '*' && keypad[ro][co] != '#' )
         {
             totalCount += getCountUtil(keypad, ro, co, n - 1 );
         }
     }
     return totalCount;
}
  
//Return count of all possible numbers of length n
//in a given numeric keyboard
static int getCount( char keypad[][], int n)
{
     //Base cases
     if (keypad == null || n <= 0 )
         return 0 ;
     if (n == 1 )
         return 10 ;
  
     int i = 0 , j = 0 , totalCount = 0 ;
     for (i = 0 ; i <4 ; i++) //Loop on keypad row
     {
         for (j= 0 ; j<3 ; j++) //Loop on keypad column
         {
             //Process for 0 to 9 digits
             if (keypad[i][j] != '*' && keypad[i][j] != '#' )
             {
                 //Get count when number is starting from key
                 //position (i, j) and add in count obtained so far
                 totalCount += getCountUtil(keypad, i, j, n);
             }
         }
     }
     return totalCount;
}
  
//Driver code
public static void main(String[] args) 
{
     char keypad[][] = {{ '1' , '2' , '3' }, { '4' , '5' , '6' }, { '7' , '8' , '9' }, { '*' , '0' , '#' }};
     System.out.printf( "Count for numbers of" +
                     " length %d: %d" , 1 , getCount(keypad, 1 ));
     System.out.printf( "\nCount for numbers of" + 
                     "length %d: %d" , 2 , getCount(keypad, 2 ));
     System.out.printf( "\nCount for numbers of" + 
                     "length %d: %d" , 3 , getCount(keypad, 3 ));
     System.out.printf( "\nCount for numbers of" + 
                     "length %d: %d" , 4 , getCount(keypad, 4 ));
     System.out.printf( "\nCount for numbers of" + 
                     "length %d: %d" , 5 , getCount(keypad, 5 ));
}
}
  
//This code has been contributed by 29AjayKumar

C#

//A Naive Recursive C# program 
//to count number of possible 
//numbers of given length
using System;
  
class GfG
{
  
//left, up, right, down 
//move from current location
static int []row = {0, 0, -1, 0, 1};
static int []col = {0, -1, 0, 1, 0};
  
//Returns count of numbers of length
//n starting from key position
//(i, j) in a numeric keyboard.
static int getCountUtil( char [, ]keypad, int i, int j, int n)
{
     if (keypad == null || n <= 0)
         return 0;
  
     //From a given key, only one 
     //number is possible of length 1
     if (n == 1)
         return 1;
  
     int k = 0, move = 0, ro = 0, co = 0, totalCount = 0;
  
     //move left, up, right, down
     //from current location and if
     //new location is valid, then 
     //get number count of length
     //(n-1) from that new position 
     //and add in count obtained so far
     for (move=0; move<5; move++)
     {
         ro = i + row[move];
         co = j + col[move];
         if (ro>= 0 && ro <= 3 && co>=0 && co <= 2 &&
         keypad[ro, co] != '*' && keypad[ro, co] != '#' )
         {
             totalCount += getCountUtil(keypad, ro, co, n - 1);
         }
     }
     return totalCount;
}
  
//Return count of all possible numbers of length n
//in a given numeric keyboard
static int getCount( char [, ]keypad, int n)
{
     //Base cases
     if (keypad == null || n <= 0)
         return 0;
     if (n == 1)
         return 10;
  
     int i = 0, j = 0, totalCount = 0;
     for (i = 0; i <4; i++) //Loop on keypad row
     {
         for (j = 0; j <3; j++) //Loop on keypad column
         {
             //Process for 0 to 9 digits
             if (keypad[i, j] != '*' && keypad[i, j] != '#' )
             {
                 //Get count when number is starting from key
                 //position (i, j) and add in count obtained so far
                 totalCount += getCountUtil(keypad, i, j, n);
             }
         }
     }
     return totalCount;
}
  
//Driver code
public static void Main() 
{
     char [, ]keypad = {{ '1' , '2' , '3' }, { '4' , '5' , '6' }, { '7' , '8' , '9' }, { '*' , '0' , '#' }};
     Console.Write( "Count for numbers of" +
                     " length {0}: {1}" , 1, getCount(keypad, 1));
     Console.Write( "\nCount for numbers of" + 
                     "length {0}: {1}" , 2, getCount(keypad, 2));
     Console.Write( "\nCount for numbers of" + 
                     "length {0}: {1}" , 3, getCount(keypad, 3));
     Console.Write( "\nCount for numbers of" + 
                     "length {0}: {1}" , 4, getCount(keypad, 4));
     Console.Write( "\nCount for numbers of" + 
                     "length {0}: {1}" , 5, getCount(keypad, 5));
}
}
  
/* This code contributed by PrinciRaj1992 */

的PHP

<?php
//A Naive Recursive PHP program 
//to count number of possible 
//numbers of given length
//left, up, right, down 
//move from current location
//static $row = array(0, 0, -1, 0, 1);
//static $col = array(0, -1, 0, 1, 0);
  
//Returns count of numbers of length
//n starting from key position
//(i, j) in a numeric keyboard.
function getCountUtil( $keypad , $i , $j , $n )
{
     static $row = array (0, 0, -1, 0, 1);
     static $col = array (0, -1, 0, 1, 0);
     if ( $keypad == null || $n <= 0)
         return 0;
  
     //From a given key, only one 
     //number is possible of length 1
     if ( $n == 1)
         return 1;
  
     $k = 0; $move = 0; $ro = 0; $co = 0; $totalCount = 0;
  
     //move left, up, right, down
     //from current location and if
     //new location is valid, then 
     //get number count of length
     //(n-1) from that new position 
     //and add in count obtained so far
     for ( $move = 0; $move <5; $move ++)
     {
         $ro = $i + $row [ $move ];
         $co = $j + $col [ $move ];
         if ( $ro>= 0 && $ro <= 3 && $co>=0 && $co <= 2 &&
         $keypad [ $ro ][ $co ] != '*' && $keypad [ $ro ][ $co ] != '#' )
         {
             $totalCount += getCountUtil( $keypad , $ro , $co , $n - 1);
         }
     }
     return $totalCount ;
}
  
//Return count of all possible numbers of length n
//in a given numeric keyboard
function getCount( $keypad , $n )
{
     //Base cases
     if ( $keypad == null || $n <= 0)
         return 0;
     if ( $n == 1)
         return 10;
  
     $i = 0; $j = 0; $totalCount = 0;
     for ( $i = 0; $i <4; $i ++) //Loop on keypad row
     {
         for ( $j = 0; $j <3; $j ++) //Loop on keypad column
         {
             //Process for 0 to 9 digits
             if ( $keypad [ $i ][ $j ] != '*' && $keypad [ $i ][ $j ] != '#' )
             {
                 //Get count when number is starting from key
                 //position (i, j) and add in count obtained so far
                 $totalCount += getCountUtil( $keypad , $i , $j , $n );
             }
         }
     }
     return $totalCount ;
}
  
//Driver code
{
     $keypad = array ( array ( '1' , '2' , '3' ), array ( '4' , '5' , '6' ), array ( '7' , '8' , '9' ), array ( '*' , '0' , '#' ));
     echo ( "Count for numbers of" . " length" . getCount( $keypad , 1)); 
     echo ( "\nCount for numbers of" . 
                     " length " . getCount( $keypad , 2));
     echo ( "\nCount for numbers of" . 
                     " length " .getCount( $keypad , 3));
     echo ( "\nCount for numbers of" .
                     " length " .getCount( $keypad , 4));
     echo ( "\nCount for numbers of" . 
                     " length " .getCount( $keypad , 5));
}
  
//This code has been contributed by Code_Mech.

输出如下：

Count for numbers of length 1: 10
Count for numbers of length 2: 36
Count for numbers of length 3: 138
Count for numbers of length 4: 532
Count for numbers of length 5: 2062

动态编程

在较小的路径上有许多重复遍历(对于较小的N而言遍历)以找到所有可能的较长的路径(对于较大的N而言遍历)。例如, 请参见以下两个图。在此遍历中, 对于从两个起始位置(按钮" 4"和" 8")开始的N = 4, 我们可以看到对于N = 2的重复遍历很少(例如4-> 1, 6-> 3, 8-> 9、8-> 7等)。

手机2
手机3

由于问题具有两个属性：最佳子结构和重叠子问题, 可以使用动态编程有效解决。

以下是用于动态编程实现的程序。

C ++

//A Dynamic Programming based C program to count number of
//possible numbers of given length
#include <stdio.h>
  
//Return count of all possible numbers of length n
//in a given numeric keyboard
int getCount( char keypad[][3], int n)
{
     if (keypad == NULL || n <= 0)
         return 0;
     if (n == 1)
         return 10;
  
     //left, up, right, down move from current location
     int row[] = {0, 0, -1, 0, 1};
     int col[] = {0, -1, 0, 1, 0};
  
     //taking n+1 for simplicity - count[i][j] will store
     //number count starting with digit i and length j
     int count[10][n+1];
     int i=0, j=0, k=0, move=0, ro=0, co=0, num = 0;
     int nextNum=0, totalCount = 0;
  
     //count numbers starting with digit i and of lengths 0 and 1
     for (i=0; i<=9; i++)
     {
         count[i][0] = 0;
         count[i][1] = 1;
     }
  
     //Bottom up - Get number count of length 2, 3, 4, ... , n
     for (k=2; k<=n; k++)
     {
         for (i=0; i<4; i++)  //Loop on keypad row
         {
             for (j=0; j<3; j++)   //Loop on keypad column
             {
                 //Process for 0 to 9 digits
                 if (keypad[i][j] != '*' && keypad[i][j] != '#' )
                 {
                     //Here we are counting the numbers starting with
                     //digit keypad[i][j] and of length k keypad[i][j]
                     //will become 1st digit, and we need to look for
                     //(k-1) more digits
                     num = keypad[i][j] - '0' ;
                     count[num][k] = 0;
  
                     //move left, up, right, down from current location
                     //and if new location is valid, then get number
                     //count of length (k-1) from that new digit and
                     //add in count we found so far
                     for (move=0; move<5; move++)
                     {
                         ro = i + row[move];
                         co = j + col[move];
                         if (ro>= 0 && ro <= 3 && co>=0 && co <= 2 &&
                            keypad[ro][co] != '*' && keypad[ro][co] != '#' )
                         {
                             nextNum = keypad[ro][co] - '0' ;
                             count[num][k] += count[nextNum][k-1];
                         }
                     }
                 }
             }
         }
     }
  
     //Get count of all possible numbers of length "n" starting
     //with digit 0, 1, 2, ..., 9
     totalCount = 0;
     for (i=0; i<=9; i++)
         totalCount += count[i][n];
     return totalCount;
}
  
//Driver program to test above function
int main( int argc, char *argv[])
{
    char keypad[4][3] = {{ '1' , '2' , '3' }, { '4' , '5' , '6' }, { '7' , '8' , '9' }, { '*' , '0' , '#' }};
    printf ( "Count for numbers of length %d: %dn" , 1, getCount(keypad, 1));
    printf ( "Count for numbers of length %d: %dn" , 2, getCount(keypad, 2));
    printf ( "Count for numbers of length %d: %dn" , 3, getCount(keypad, 3));
    printf ( "Count for numbers of length %d: %dn" , 4, getCount(keypad, 4));
    printf ( "Count for numbers of length %d: %dn" , 5, getCount(keypad, 5));
  
    return 0;
}

Java

//A Dynamic Programming based Java program to 
//count number of possible numbers of given length
class GFG
{
      
//Return count of all possible numbers of length n
//in a given numeric keyboard
static int getCount( char keypad[][], int n)
{
     if (keypad == null || n <= 0 )
         return 0 ;
     if (n == 1 )
         return 10 ;
  
     //left, up, right, down move from current location
     int row[] = { 0 , 0 , - 1 , 0 , 1 };
     int col[] = { 0 , - 1 , 0 , 1 , 0 };
  
     //taking n+1 for simplicity - count[i][j] will store
     //number count starting with digit i and length j
     int [][]count = new int [ 10 ][n + 1 ];
     int i = 0 , j = 0 , k = 0 , move = 0 , ro = 0 , co = 0 , num = 0 ;
     int nextNum = 0 , totalCount = 0 ;
  
     //count numbers starting with digit i 
     //and of lengths 0 and 1
     for (i = 0 ; i <= 9 ; i++)
     {
         count[i][ 0 ] = 0 ;
         count[i][ 1 ] = 1 ;
     }
  
     //Bottom up - Get number count of length 2, 3, 4, ... , n
     for (k = 2 ; k <= n; k++)
     {
         for (i = 0 ; i <4 ; i++) //Loop on keypad row
         {
             for (j = 0 ; j <3 ; j++) //Loop on keypad column
             {
                 //Process for 0 to 9 digits
                 if (keypad[i][j] != '*' && 
                     keypad[i][j] != '#' )
                 {
                     //Here we are counting the numbers starting with
                     //digit keypad[i][j] and of length k keypad[i][j]
                     //will become 1st digit, and we need to look for
                     //(k-1) more digits
                     num = keypad[i][j] - '0' ;
                     count[num][k] = 0 ;
  
                     //move left, up, right, down from current location
                     //and if new location is valid, then get number
                     //count of length (k-1) from that new digit and
                     //add in count we found so far
                     for (move = 0 ; move <5 ; move++)
                     {
                         ro = i + row[move];
                         co = j + col[move];
                         if (ro>= 0 && ro <= 3 && co>= 0 && 
                             co <= 2 && keypad[ro][co] != '*' && 
                                        keypad[ro][co] != '#' )
                         {
                             nextNum = keypad[ro][co] - '0' ;
                             count[num][k] += count[nextNum][k - 1 ];
                         }
                     }
                 }
             }
         }
     }
  
     //Get count of all possible numbers of length "n" 
     //starting with digit 0, 1, 2, ..., 9
     totalCount = 0 ;
     for (i = 0 ; i <= 9 ; i++)
         totalCount += count[i][n];
     return totalCount;
}
  
//Driver Code
public static void main(String[] args)
{
     char keypad[][] = {{ '1' , '2' , '3' }, { '4' , '5' , '6' }, { '7' , '8' , '9' }, { '*' , '0' , '#' }};
     System.out.printf( "Count for numbers of length %d: %d\n" , 1 , getCount(keypad, 1 ));
     System.out.printf( "Count for numbers of length %d: %d\n" , 2 , getCount(keypad, 2 ));
     System.out.printf( "Count for numbers of length %d: %d\n" , 3 , getCount(keypad, 3 ));
     System.out.printf( "Count for numbers of length %d: %d\n" , 4 , getCount(keypad, 4 ));
     System.out.printf( "Count for numbers of length %d: %d\n" , 5 , getCount(keypad, 5 ));
}
}
  
//This code is contributed by Rajput-Ji

C#

//A Dynamic Programming based C# program to 
//count number of possible numbers of given Length
using System;
  
class GFG
{
      
//Return count of all possible numbers of Length n
//in a given numeric keyboard
static int getCount( char [, ]keypad, int n)
{
     if (keypad == null || n <= 0)
         return 0;
     if (n == 1)
         return 10;
  
     //left, up, right, down move
     //from current location
     int []row = {0, 0, -1, 0, 1};
     int []col = {0, -1, 0, 1, 0};
  
     //taking n+1 for simplicity - count[i, j] 
     //will store number count starting with
     //digit i and.Length j
     int [, ]count = new int [10, n + 1];
     int i = 0, j = 0, k = 0, move = 0, ro = 0, co = 0, num = 0;
     int nextNum = 0, totalCount = 0;
  
     //count numbers starting with digit i 
     //and of.Lengths 0 and 1
     for (i = 0; i <= 9; i++)
     {
         count[i, 0] = 0;
         count[i, 1] = 1;
     }
  
     //Bottom up - Get number count of
     //Length 2, 3, 4, ... , n
     for (k = 2; k <= n; k++)
     {
         for (i = 0; i <4; i++) //Loop on keypad row
         {
             for (j = 0; j <3; j++) //Loop on keypad column
             {
                 //Process for 0 to 9 digits
                 if (keypad[i, j] != '*' && 
                     keypad[i, j] != '#' )
                 {
                     //Here we are counting the numbers starting with
                     //digit keypad[i, j] and of.Length k keypad[i, j]
                     //will become 1st digit, and we need to look for
                     //(k-1) more digits
                     num = keypad[i, j] - '0' ;
                     count[num, k] = 0;
  
                     //move left, up, right, down from current location
                     //and if new location is valid, then get number
                     //count of.Length (k-1) from that new digit and
                     //.Add in count we found so far
                     for (move = 0; move <5; move++)
                     {
                         ro = i + row[move];
                         co = j + col[move];
                         if (ro>= 0 && ro <= 3 && co>= 0 && 
                             co <= 2 && keypad[ro, co] != '*' && 
                                        keypad[ro, co] != '#' )
                         {
                             nextNum = keypad[ro, co] - '0' ;
                             count[num, k] += count[nextNum, k - 1];
                         }
                     }
                 }
             }
         }
     }
  
     //Get count of all possible numbers of.Length "n" 
     //starting with digit 0, 1, 2, ..., 9
     totalCount = 0;
     for (i = 0; i <= 9; i++)
         totalCount += count[i, n];
     return totalCount;
}
  
//Driver Code
public static void Main(String[] args)
{
     char [, ]keypad = {{ '1' , '2' , '3' }, { '4' , '5' , '6' }, { '7' , '8' , '9' }, { '*' , '0' , '#' }};
     Console.Write( "Count for numbers of.Length {0}: {1}\n" , 1, getCount(keypad, 1));
     Console.Write( "Count for numbers of.Length {0}: {1}\n" , 2, getCount(keypad, 2));
     Console.Write( "Count for numbers of.Length {0}: {1}\n" , 3, getCount(keypad, 3));
     Console.Write( "Count for numbers of.Length {0}: {1}\n" , 4, getCount(keypad, 4));
     Console.Write( "Count for numbers of.Length {0}: {1}\n" , 5, getCount(keypad, 5));
}
}
  
//This code is contributed by Rajput-Ji

输出如下：

Count for numbers of length 1: 10
Count for numbers of length 2: 36
Count for numbers of length 3: 138
Count for numbers of length 4: 532
Count for numbers of length 5: 2062

空间优化的解决方案：

上述动态编程方法也需要O(n)时间运行, 并且需要O(n)辅助空间, 因为只有一个for循环运行n次, 其他for循环运行恒定时间。我们可以看到第n次迭代仅需要第(n-1)次迭代中的数据, 因此我们不需要保留较旧迭代中的数据。我们可以使用只有两个大小为10的数组的高效空间动态编程方法。感谢Nik提出了这种解决方案。

C ++

//A Space Optimized C program to count number of possible numbers
//of given length
#include <stdio.h>
  
//Return count of all possible numbers of length n
//in a given numeric keyboard
int getCount( char keypad[][3], int n)
{
     if (keypad == NULL || n <= 0)
         return 0;
     if (n == 1)
         return 10;
  
     //odd[i], even[i] arrays represent count of numbers starting
     //with digit i for any length j
     int odd[10], even[10];
     int i = 0, j = 0, useOdd = 0, totalCount = 0;
  
     for (i=0; i<=9; i++)
         odd[i] = 1;  //for j = 1
  
     for (j=2; j<=n; j++) //Bottom Up calculation from j = 2 to n
     {
         useOdd = 1 - useOdd;
  
         //Here we are explicitly writing lines for each number 0
         //to 9. But it can always be written as DFS on 4X3 grid
         //using row, column array valid moves
         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];
         }
     }
  
     //Get count of all possible numbers of length "n" starting
     //with digit 0, 1, 2, ..., 9
     totalCount = 0;
     if (useOdd == 1)
     {
         for (i=0; i<=9; i++)
             totalCount += even[i];
     }
     else
     {
         for (i=0; i<=9; i++)
             totalCount += odd[i];
     }
     return totalCount;
}
  
//Driver program to test above function
int main()
{
     char keypad[4][3] = {{ '1' , '2' , '3' }, { '4' , '5' , '6' }, { '7' , '8' , '9' }, { '*' , '0' , '#' }
     };
     printf ( "Count for numbers of length %d: %dn" , 1, getCount(keypad, 1));
     printf ( "Count for numbers of length %d: %dn" , 2, getCount(keypad, 2));
     printf ( "Count for numbers of length %d: %dn" , 3, getCount(keypad, 3));
     printf ( "Count for numbers of length %d: %dn" , 4, getCount(keypad, 4));
     printf ( "Count for numbers of length %d: %dn" , 5, getCount(keypad, 5));
  
     return 0;
}

Java

//A Space Optimized Java program to 
//count number of possible numbers 
//of given length
class GFG 
{
  
//Return count of all possible numbers of 
//length n in a given numeric keyboard
static int getCount( char keypad[][], int n)
{
     if (keypad == null || n <= 0 )
         return 0 ;
     if (n == 1 )
         return 10 ;
  
     //odd[i], even[i] arrays represent count of 
     //numbers starting with digit i for any length j
     int []odd = new int [ 10 ];
     int []even = new int [ 10 ];
     int i = 0 , j = 0 , useOdd = 0 , totalCount = 0 ;
  
     for (i = 0 ; i <= 9 ; i++)
         odd[i] = 1 ; //for j = 1
      
     //Bottom Up calculation from j = 2 to n
     for (j = 2 ; j <= n; j++) 
     {
         useOdd = 1 - useOdd;
  
         //Here we are explicitly writing lines 
         //for each number 0 to 9. But it can always be 
         //written as DFS on 4X3 grid using row, //column array valid moves
         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 ];
         }
     }
  
     //Get count of all possible numbers of 
     //length "n" starting with digit 0, 1, 2, ..., 9
     totalCount = 0 ;
     if (useOdd == 1 )
     {
         for (i = 0 ; i <= 9 ; i++)
             totalCount += even[i];
     }
     else
     {
         for (i = 0 ; i <= 9 ; i++)
             totalCount += odd[i];
     }
     return totalCount;
}
  
//Driver Code
public static void main(String[] args) 
{
     char keypad[][] = {{ '1' , '2' , '3' }, { '4' , '5' , '6' }, { '7' , '8' , '9' }, { '*' , '0' , '#' }};
     System.out.printf( "Count for numbers of length %d: %d\n" , 1 , getCount(keypad, 1 ));
     System.out.printf( "Count for numbers of length %d: %d\n" , 2 , getCount(keypad, 2 ));
     System.out.printf( "Count for numbers of length %d: %d\n" , 3 , getCount(keypad, 3 ));
     System.out.printf( "Count for numbers of length %d: %d\n" , 4 , getCount(keypad, 4 ));
     System.out.printf( "Count for numbers of length %d: %d\n" , 5 , getCount(keypad, 5 ));
}
}
  
//This code is contributed by PrinciRaj1992

Python 3

# A Space Optimized Python program to count 
# number of possible numbers
# of given length
  
# Return count of all possible numbers 
# of length n
# in a given numeric keyboard
def getCount(keypad, n):
  
     if ( not keypad or n <= 0 ):
         return 0
     if (n = = 1 ):
         return 10
  
     # odd[i], even[i] arrays represent 
     # count of numbers starting
     # with digit i for any length j
     odd = [ 0 ] * 10
     even = [ 0 ] * 10
     i = 0
     j = 0
     useOdd = 0
     totalCount = 0
  
     for i in range ( 10 ):
         odd[i] = 1 # for j = 1
  
     for j in range ( 2 , n + 1 ): # Bottom Up calculation from j = 2 to n
      
         useOdd = 1 - useOdd
  
         # Here we are explicitly writing lines for each number 0
         # to 9. But it can always be written as DFS on 4X3 grid
         # using row, column array valid moves
         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 ]
  
     # Get count of all possible numbers of length "n" starting
     # with digit 0, 1, 2, ..., 9
     totalCount = 0
     if (useOdd = = 1 ):
         for i in range ( 10 ):
             totalCount + = even[i]
      
     else :
         for i in range ( 10 ):
             totalCount + = odd[i]
  
     return totalCount
  
# Driver program to test above function
if __name__ = = "__main__" :
     keypad = [[ '1' , '2' , '3' ], [ '4' , '5' , '6' ], [ '7' , '8' , '9' ], [ '*' , '0' , '#' ]]
      
     print ( "Count for numbers of length " , 1 , ": " , getCount(keypad, 1 ))
     print ( "Count for numbers of length " , 2 , ": " , getCount(keypad, 2 ))
     print ( "Count for numbers of length " , 3 , ": " , getCount(keypad, 3 ))
     print ( "Count for numbers of length " , 4 , ": " , getCount(keypad, 4 ))
     print ( "Count for numbers of length " , 5 , ": " , getCount(keypad, 5 ))
      
# This code is contributed by
# ChitraNayal

C#

//A Space Optimized C# program to 
//count number of possible numbers 
//of given length
using System;
      
class GFG 
{
  
//Return count of all possible numbers of 
//length n in a given numeric keyboard
static int getCount( char [, ]keypad, int n)
{
     if (keypad == null || n <= 0)
         return 0;
     if (n == 1)
         return 10;
  
     //odd[i], even[i] arrays represent count of 
     //numbers starting with digit i for any length j
     int []odd = new int [10];
     int []even = new int [10];
     int i = 0, j = 0, useOdd = 0, totalCount = 0;
  
     for (i = 0; i <= 9; i++)
         odd[i] = 1; //for j = 1
      
     //Bottom Up calculation from j = 2 to n
     for (j = 2; j <= n; j++) 
     {
         useOdd = 1 - useOdd;
  
         //Here we are explicitly writing lines 
         //for each number 0 to 9. But it can always be 
         //written as DFS on 4X3 grid using row, //column array valid moves
         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];
         }
     }
  
     //Get count of all possible numbers of 
     //length "n" starting with digit 0, 1, 2, ..., 9
     totalCount = 0;
     if (useOdd == 1)
     {
         for (i = 0; i <= 9; i++)
             totalCount += even[i];
     }
     else
     {
         for (i = 0; i <= 9; i++)
             totalCount += odd[i];
     }
     return totalCount;
}
  
//Driver Code
public static void Main(String[] args) 
{
     char [, ]keypad = {{ '1' , '2' , '3' }, { '4' , '5' , '6' }, { '7' , '8' , '9' }, { '*' , '0' , '#' }};
     Console.Write( "Count for numbers of length {0}: {1}\n" , 1, getCount(keypad, 1));
     Console.Write( "Count for numbers of length {0}: {1}\n" , 2, getCount(keypad, 2));
     Console.Write( "Count for numbers of length {0}: {1}\n" , 3, getCount(keypad, 3));
     Console.Write( "Count for numbers of length {0}: {1}\n" , 4, getCount(keypad, 4));
     Console.Write( "Count for numbers of length {0}: {1}\n" , 5, getCount(keypad, 5));
}
}
  
//This code is contributed by 29AjayKumar

输出如下：

Count for numbers of length 1: 10
Count for numbers of length 2: 36
Count for numbers of length 3: 138
Count for numbers of length 4: 532
Count for numbers of length 5: 2062

本文作者：阿努拉格·辛格(Anurag Singh)。如果发现任何不正确的地方, 或者想分享有关上述主题的更多信息, 请写评论。

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