本文概述

• C ++
• Java
• C#
• 的PHP
• C ++
• Java
• C#
• C ++
• Java
• Python 3
• C#

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＃)。

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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)。

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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 = 1for (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 = 0for i in range ( 10 ): odd[i] = 1 # for j = 1for j in range ( 2 , n + 1 ): # Bottom Up calculation from j = 2 to nuseOdd = 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

【手机数字键盘问题详细介绍】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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