相邻数字之间具有绝对差异的N位数字的计数为K

给定两个整数N和K。任务是计算所有长度为 N且相邻数字之间的绝对差等于K的正整数。

例子：

Input: N = 4, K = 8
Output: 3
Explanation: The absolute difference between every consecutive digit of each number is 8. Three possible numbers are 8080, 1919 and 9191.

Input: N = 2, K = 0
Output: 9
Explanation: 11, 22, 33, 44, 55, 66, 77, 88, 99. The absolute difference between every consecutive digit of each number is 0.

编程需要懂一点英语

方法：该方法基于递归。迭代数字[1, 9]，对于每个数字，使用递归将具有绝对数字差异的N位数字计数为K。以下情况到达递归函数调用。

基本情况：对于所有个位数整数，即 N = 1，增加答案计数。
递归调用：如果到现在形成的数字(num)的个位数加上数字K不超过9，则通过减少N并使得num = (num*10 + num%10 + K)递归调用。

if(num % 10 + K ≤ 9)
recursive_function(10 * num + (num % 10 + K), N – 1);

编程需要懂一点英语

如果在所有递归调用之后K的值非零并且如果num % 10 ≥ K ，则通过减小N并将num更新为(10*num + num%10 – K) 再次递归调用。

if(num % 10 ≥ K)
recursive_function(10 * num + num % 10 – K, N – 1)

编程需要懂一点英语

下面是上述方法的实现。

C++

// C++ program for the above approach
#include 
using namespace std;
 
// To store the count of numbers
int countNums = 0;
 
// Function that recursively finds the
// possible numbers and append into ans
void checkUtil(int num, int K, int N)
{
    // Base Case
    if (N == 1) {
        countNums++;
        return;
    }
 
    // Check the sum of last digit and k
    // less than or equal to 9 or not
    if ((num % 10 + K) <= 9)
        checkUtil(10 * num +
                  (num % 10 + K), K, N - 1);
 
    // If K = 0, then subtraction and
    // addition does not make any
    // difference
    if (K) {
         
        // If unit's digit greater than K
        if ((num % 10 - K) >= 0)
            checkUtil(10 * num +
                      num % 10 - K, K, N - 1);
    }
}
 
// Function to call checkUtil function
// for every integer from 1 to 9
void check(int K, int N)
{
    // Loop to check for
    // all digits from 1 to 9
    for (int i = 1; i <= 9; i++) {
        checkUtil(i, K, N);
    }
}
 
// Driver Code
int main()
{
    // Given N and K
    int N = 4, K = 8;
    check(K, N);
 
    // Count total possible numbers
    cout << countNums << endl;
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
// To store the count of numbers
static int countNums = 0;
 
// Function that recursively finds the
// possible numbers and append into ans
static void checkUtil(int num, int K, int N)
{
   
    // Base Case
    if (N == 1) {
        countNums++;
        return;
    }
 
    // Check the sum of last digit and k
    // less than or equal to 9 or not
    if ((num % 10 + K) <= 9)
        checkUtil(10 * num +
                  (num % 10 + K), K, N - 1);
 
    // If K = 0, then subtraction and
    // addition does not make any
    // difference
    if (K>0) {
         
        // If unit's digit greater than K
        if ((num % 10 - K) >= 0)
            checkUtil(10 * num +
                      num % 10 - K, K, N - 1);
    }
}
 
// Function to call checkUtil function
// for every integer from 1 to 9
static void check(int K, int N)
{
   
    // Loop to check for
    // all digits from 1 to 9
    for (int i = 1; i <= 9; i++) {
        checkUtil(i, K, N);
    }
}
 
// Driver Code
public static void main(String[] args)
{
   
    // Given N and K
    int N = 4, K = 8;
    check(K, N);
 
    // Count total possible numbers
    System.out.print(countNums +"\n");
}
}
 
// This code contributed by shikhasingrajput

Python3

# Python program for the above approach
 
# To store the count of numbers
countNums = 0;
 
# Function that recursively finds the
# possible numbers and append into ans
def checkUtil(num, K, N):
    global countNums;
     
    # Base Case
    if (N == 1):
        countNums += 1;
        return;
 
    # Check the sum of last digit and k
    # less than or equal to 9 or not
    if ((num % 10 + K) <= 9):
        checkUtil(10 * num + (num % 10 + K), K, N - 1);
 
    # If K = 0, then subtraction and
    # addition does not make any
    # difference
    if (K > 0):
 
        # If unit's digit greater than K
        if ((num % 10 - K) >= 0):
            checkUtil(10 * num + num % 10 - K, K, N - 1);
 
# Function to call checkUtil function
# for every integer from 1 to 9
def check(K, N):
   
    # Loop to check for
    # all digits from 1 to 9
    for i in range(1,10):
        checkUtil(i, K, N);
 
# Driver Code
if __name__ == '__main__':
    # Given N and K
    N = 4;
    K = 8;
    check(K, N);
 
    # Count total possible numbers
    print(countNums);
 
# This code is contributed by shikhasingrajput

C#

// C# program for the above approach
using System;
 
class GFG{
 
// To store the count of numbers
static int countNums = 0;
 
// Function that recursively finds the
// possible numbers and append into ans
static void checkUtil(int num, int K, int N)
{
     
    // Base Case
    if (N == 1)
    {
        countNums++;
        return;
    }
 
    // Check the sum of last digit and k
    // less than or equal to 9 or not
    if ((num % 10 + K) <= 9)
        checkUtil(10 * num +
                 (num % 10 + K), K, N - 1);
 
    // If K = 0, then subtraction and
    // addition does not make any
    // difference
    if (K > 0)
    {
         
        // If unit's digit greater than K
        if ((num % 10 - K) >= 0)
            checkUtil(10 * num +
                      num % 10 - K, K, N - 1);
    }
}
 
// Function to call checkUtil function
// for every integer from 1 to 9
static void check(int K, int N)
{
     
    // Loop to check for
    // all digits from 1 to 9
    for(int i = 1; i <= 9; i++)
    {
        checkUtil(i, K, N);
    }
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given N and K
    int N = 4, K = 8;
    check(K, N);
 
    // Count total possible numbers
    Console.Write(countNums + "\n");
}
}
 
// This code is contributed by 29AjayKumar

Javascript

输出

时间复杂度： O(2 ^N )
辅助空间： O(1)