n 位步进数的个数 |空间优化解决方案

给定 n，找出 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
The numbers are 10, 12, 21, 
23, 32, 34, 43, 45, 54, 56, 65, 67, 76, 
78, 87, 89, 98.

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

在上一篇文章中，讨论了一个需要 O(n) 辅助空间的解决方案。可以优化解决问题所需的辅助空间。二维 dp 数组 dp[i][j] 表示长度为 i 和最后一位为 j 的步进数的计数。对于数字 j，从数字 j – 1 和 j + 1 中获得计数。递推关系为dp[i][j] = dp[i-1][j-1] + dp[i-1][j +1] 。观察当前长度 i 的答案仅取决于 i – 1。因此可以使用一维 dp 数组，其中对于给定的 i，dp[j] 存储长度 i 以数字 j 结尾的步进数的计数。在更新给定长度 i 的 dp 数组之前，将长度 i – 1 的结果存储在另一个数组prev 中，然后使用prev数组更新 dp 数组。
下面是上述方法的实现：

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[j] stores count of i digit
    // stepping numbers ending with digit
    // j.
    int dp[10];
 
    // To store result of length i - 1
    // before updating dp[j] for length i.
    int prev[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[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++) {
            prev[j] = dp[j];
        }
 
        for (int j = 0; j <= 9; j++) {
 
            // If ending digit is 0
            if (j == 0)
                dp[j] = prev[j + 1];
 
            // If ending digit is 9
            else if (j == 9)
                dp[j] = prev[j - 1];
 
            // For other digits.
            else
                dp[j] = prev[j - 1] + prev[j + 1];
        }
    }
 
    // stores the final answer
    long long sum = 0;
    for (int j = 1; j <= 9; j++)
        sum += dp[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[j] stores count of i digit
    // stepping numbers ending with digit
    // j.
    int[] dp = new int[10];
 
    // To store result of length i - 1
    // before updating dp[j] for length i.
    int[] prev = new int[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[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++)
        {
            prev[j] = dp[j];
        }
 
        for (int j = 0; j <= 9; j++)
        {
 
            // If ending digit is 0
            if (j == 0)
                dp[j] = prev[j + 1];
 
            // If ending digit is 9
            else if (j == 9)
                dp[j] = prev[j - 1];
 
            // For other digits.
            else
                dp[j] = prev[j - 1] + prev[j + 1];
        }
    }
 
    // stores the final answer
    long sum = 0;
    for (int j = 1; j <= 9; j++)
        sum += dp[j];
    return sum;
}
 
// Driver code
public static void main (String[] args)
{
    int n = 2;
    System.out.println(answer(n));
}
}
 
// This code is contributed by mits

Python3

# Python3 program to calculate the number of
# n digit stepping numbers.
 
# function that calculates the answer
def answer(n) :
 
    # dp[j] stores count of i digit
    # stepping numbers ending with digit j.
    dp = [0] * 10
 
    # To store resu1lt of length i - 1
    # before updating dp[j] for length i.
    prev = [0] * 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 j in range(0, 10) :
        dp[j] = 1
 
    # Compute values for count of digits
    # more than 1.
    for i in range(2, n + 1):
        for j in range (0, 10):
            prev[j] = dp[j]
 
        for j in range (0, 10):
         
            # If ending digit is 0
            if (j == 0):
                dp[j] = prev[j + 1]
 
            # If ending digit is 9
            elif (j == 9) :
                dp[j] = prev[j - 1]
 
            # For other digits.
            else :
                dp[j] = prev[j - 1] + prev[j + 1]
         
    # stores the final answer
    sum = 0
    for j in range (1, 10):
        sum = sum + dp[j]
    return sum
 
# Driver Code
n = 2
print(answer(n))
 
# This code is contributed by ihritik

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[j] stores count of i digit
    // stepping numbers ending with digit
    // j.
    int[] dp = new int[10];
 
    // To store result of length i - 1
    // before updating dp[j] for length i.
    int[] prev = new int[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[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++)
        {
            prev[j] = dp[j];
        }
 
        for (int j = 0; j <= 9; j++)
        {
 
            // If ending digit is 0
            if (j == 0)
                dp[j] = prev[j + 1];
 
            // If ending digit is 9
            else if (j == 9)
                dp[j] = prev[j - 1];
 
            // For other digits.
            else
                dp[j] = prev[j - 1] + prev[j + 1];
        }
    }
 
    // stores the final answer
    long sum = 0;
    for (int j = 1; j <= 9; j++)
        sum += dp[j];
    return sum;
}
 
// Driver code
static void Main()
{
    int n = 2;
    Console.WriteLine(answer(n));
}
}
 
// This code is contributed by mits

Javascript

 
 
// Javascript program to calculate the number of
// n digit stepping numbers.
 
// function that calculates the answer
function answer(n)
{
    // dp[j] stores count of i digit
    // stepping numbers ending with digit
    // j.
    var dp = Array(10);
 
    // To store result of length i - 1
    // before updating dp[j] for length i.
    var prev = Array(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 (var j = 0; j <= 9; j++)
        dp[j] = 1;
 
    // Compute values for count of digits
    // more than 1.
    for (var i = 2; i <= n; i++) {
        for (var j = 0; j <= 9; j++) {
            prev[j] = dp[j];
        }
 
        for (var j = 0; j <= 9; j++) {
 
            // If ending digit is 0
            if (j == 0)
                dp[j] = prev[j + 1];
 
            // If ending digit is 9
            else if (j == 9)
                dp[j] = prev[j - 1];
 
            // For other digits.
            else
                dp[j] = prev[j - 1] + prev[j + 1];
        }
    }
 
    // stores the final answer
    var sum = 0;
    for (var j = 1; j <= 9; j++)
        sum += dp[j];
    return sum;
}
 
// driver program to test the above function
var n = 2;
document.write( answer(n));
 

输出：

17

时间复杂度： O(N)
辅助空间： O(1)

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