n 位步进数

给定 n，找到 n 位步进数的计数。如果所有相邻数字的绝对差为 1，则该数字称为步进数。321 是步进数，而 421 不是。

例子 ：

Input : 2 
Output : 17
Explanation: The numbers are 10, 12, 21, 
23, 32, 34, 43, 45, 54, 56, 65, 67, 76, 
78, 87, 89, 98.

Input : 1
Output : 10
Explanation: the numbers are 0, 1, 2, 3, 
4, 5, 6, 7, 8, 9.

一种天真的方法是对所有 n 位数字运行一个循环，并检查每个数字是否为步进。

一种有效的方法是使用动态规划。

In dp[i][j], i denotes number of
digits and j denotes last digit.

// If there is only one digit
if (i == 1)
   dp(i, j) = 1;

// If last digit is 0.
if (j == 0) 
   dp(i, j) = dp(i-1, j+1)

// If last digit is 9
else if (j == 9) 
   dp(i, j) = dp(i-1, j-1)

// If last digit is neither 0
// nor 9.
else 
   dp(i, j) = dp(i-1, j-1) + 
              dp(i-1, j+1)

Result is ∑dp(n, j) where j varies
from 1 to 9. 

C++

// CPP program to calculate the number of
// n digit stepping numbers.
#include 
using namespace std;
 
// function that calculates the answer
long long answer(int n)
{
    // dp[i][j] stores count of i digit
    // stepping numbers ending with digit
    // j.
    int dp[n + 1][10];
 
    // if n is 1 then answer will be 10.
    if (n == 1)
        return 10;
 
    // Initialize values for count of
    // digits equal to 1.
    for (int j = 0; j <= 9; j++)
        dp[1][j] = 1;
 
    // Compute values for count of digits
    // more than 1.
    for (int i = 2; i <= n; i++) {
        for (int j = 0; j <= 9; j++) {
 
            // If ending digit is 0
            if (j == 0)
                dp[i][j] = dp[i - 1][j + 1];
 
            // If ending digit is 9
            else if (j == 9)
                dp[i][j] = dp[i - 1][j - 1];
 
            // For other digits.
            else
                dp[i][j] = dp[i - 1][j - 1] +
                           dp[i - 1][j + 1];
        }
    }
 
    // stores the final answer
    long long sum = 0;
    for (int j = 1; j <= 9; j++)
        sum += dp[n][j];
    return sum;
}
 
// driver program to test the above function
int main()
{
    int n = 2;
    cout << answer(n);
    return 0;
}

Java

// Java program to calculate the number of
// n digit stepping numbers.
 
class GFG {
         
    // function that calculates the answer
    static long answer(int n)
    {
        // dp[i][j] stores count of i
        // digit stepping numbers ending
        // with digit j.
        int dp[][] = new int[n+1][10];
      
        // if n is 1 then answer will be 10.
        if (n == 1)
            return 10;
      
        // Initialize values for count of
        // digits equal to 1.
        for (int j = 0; j <= 9; j++)
            dp[1][j] = 1;
      
        // Compute values for count of
        // digits more than 1.
        for (int i = 2; i <= n; i++) {
            for (int j = 0; j <= 9; j++) {
      
                // If ending digit is 0
                if (j == 0)
                    dp[i][j] = dp[i - 1][j + 1];
      
                // If ending digit is 9
                else if (j == 9)
                    dp[i][j] = dp[i - 1][j - 1];
      
                // For other digits.
                else
                    dp[i][j] = dp[i - 1][j - 1] +
                               dp[i - 1][j + 1];
            }
        }
      
        // stores the final answer
        long sum = 0;
        for (int j = 1; j <= 9; j++)
            sum += dp[n][j];
        return sum;
    }
      
    // driver program to test the above function
    public static void main(String args[])
    {
        int n = 2;
        System.out.println(answer(n));
    }
}
 
/*This code is contributed by Nikita tiwari.*/

Python3

# Python3 program to calculate
# the number of n digit
# stepping numbers.
 
# function that calculates
# the answer
def answer(n):
     
    # dp[i][j] stores count of
    # i digit stepping numbers
    # ending with digit j.
    dp = [[0 for x in range(10)]
             for y in range(n + 1)];
 
    # if n is 1 then answer
    # will be 10.
    if (n == 1):
        return 10;
    for j in range(10):
        dp[1][j] = 1;
 
    # Compute values for count
    # of digits more than 1.
    for i in range(2, n + 1):
        for j in range(10):
             
            # If ending digit is 0
            if (j == 0):
                dp[i][j] = dp[i - 1][j + 1];
                 
            # If ending digit is 9
            elif (j == 9):
                dp[i][j] = dp[i - 1][j - 1];
                 
            # For other digits.
            else:
                dp[i][j] = (dp[i - 1][j - 1] +
                            dp[i - 1][j + 1]);
                 
    # stores the final answer
    sum = 0;
    for j in range(1, 10):
        sum = sum + dp[n][j];
    return sum;
 
# Driver Code
n = 2;
print(answer(n));
 
# This code is contributed
# by mits

C#

// C# program to calculate the number of
// n digit stepping numbers.
using System;
 
class GFG {
         
    // function that calculates the answer
    static long answer(int n)
    {
         
        // dp[i][j] stores count of i
        // digit stepping numbers ending
        // with digit j.
        int [,]dp = new int[n+1,10];
     
        // if n is 1 then answer will be 10.
        if (n == 1)
            return 10;
     
        // Initialize values for count of
        // digits equal to 1.
        for (int j = 0; j <= 9; j++)
            dp[1,j] = 1;
     
        // Compute values for count of
        // digits more than 1.
        for (int i = 2; i <= n; i++) {
            for (int j = 0; j <= 9; j++) {
     
                // If ending digit is 0
                if (j == 0)
                    dp[i,j] = dp[i - 1,j + 1];
     
                // If ending digit is 9
                else if (j == 9)
                    dp[i,j] = dp[i - 1,j - 1];
     
                // For other digits.
                else
                    dp[i,j] = dp[i - 1,j - 1] +
                               dp[i - 1,j + 1];
            }
        }
     
        // stores the final answer
        long sum = 0;
        for (int j = 1; j <= 9; j++)
            sum += dp[n,j];
             
        return sum;
    }
     
    // driver program to test the above function
    public static void Main()
    {
        int n = 2;
         
        Console.WriteLine(answer(n));
    }
}
 
/*This code is contributed by vt_m.*/

Javascript

 
// JavaScript program to calculate the number of
// n digit stepping numbers.
 
// Function that calculates the answer
function answer(n)
{
     
    // dp[i][j] stores count of i
    // digit stepping numbers ending
    // with digit j.
    let dp = new Array(n + 1);
     
    // Loop to create 2D array using 1D array
    for(var i = 0; i < dp.length; i++)
    {
        dp[i] = new Array(2);
    }
    
    // If n is 1 then answer will be 10.
    if (n == 1)
        return 10;
    
    // Initialize values for count of
    // digits equal to 1.
    for(let j = 0; j <= 9; j++)
        dp[1][j] = 1;
    
    // Compute values for count of
    // digits more than 1.
    for(let i = 2; i <= n; i++)
    {
        for(let j = 0; j <= 9; j++)
        {
    
            // If ending digit is 0
            if (j == 0)
                dp[i][j] = dp[i - 1][j + 1];
    
            // If ending digit is 9
            else if (j == 9)
                dp[i][j] = dp[i - 1][j - 1];
    
            // For other digits.
            else
                dp[i][j] = dp[i - 1][j - 1] +
                           dp[i - 1][j + 1];
        }
    }
    
    // Stores the final answer
    let sum = 0;
    for(let j = 1; j <= 9; j++)
        sum += dp[n][j];
         
    return sum;
}
   
// Driver Code
let n = 2;
 
document.write(answer(n));
 
// This code is contributed by code_hunt
 

输出 ：

17

时间复杂度： O(n)
辅助空间： O(n)n 位步进数的个数 |空间优化解决方案