程序打印前N个步进编号

给定数字N ，任务是打印前N个步进编号。

A number is called stepping number if all adjacent digits have an absolute difference of 1. For e.g. 321 is a Stepping Number while 421 is not.

例子：

Input: N = 7
Output: 1, 2, 3, 4, 5, 6, 7

Input: N = 14
Output: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 21, 22

幼稚的方法：我们可以从1开始，检查每个数字是否为Stepping编号，然后继续直到找到第K个Stepping编号。

高效方法：

1. 生成所有可能的步进数，直到1000，以便于计算
2. 对于N的每个值，只需打印已计算的N个步进数即可

下面是上述方法的实现：

C++

// C++ Program to print first N
// Stepping numbers
  
#include 
using namespace std;
  
// Function to generate
// the Stepping numbers
void generateSteppingNos(
    int x, set& s)
{
    if (x > 1e8)
        return;
  
    // Inserting the current
    // element in the set
    s.insert(x);
  
    // Retrieving the last digit
    // of the current number
    int last = x % 10;
  
    if (last - 1 >= 0)
  
        // Appending x-1 to
        // the current number
        generateSteppingNos(
            x * 10 + last - 1,
            s);
  
    // Appending x to
    // the current number
    generateSteppingNos(
        x * 10 + last,
        s);
  
    if (last + 1 <= 9)
  
        // Appending x+1 to
        // the current number
        generateSteppingNos(
            x * 10 + last + 1,
            s);
}
  
// Function to print
// N Stepping numbers
void NSteppingNumbers(int N)
{
    set s;
  
    for (int i = 1; i <= 9; i++)
        generateSteppingNos(i, s);
  
    int count = 1;
  
    // Printing N numbers from s
    for (auto& it : s) {
        if (count <= N) {
            cout << it << ", ";
            count++;
        }
        else
            break;
    }
}
  
// Driver code
int main()
{
    int N = 14;
  
    NSteppingNumbers(N);
  
    return 0;
}

Python3

# Python3 Program to print first N 
# Stepping numbers 
  
# Function to generate 
# the Stepping numbers 
def generateSteppingNos(x, s): 
    if (x > 1e8):
        return
  
    # Inserting the current 
    # element in the set 
    s.add(x)
  
    # Retrieving the last digit 
    # of the current number 
    last = x % 10
  
    if (last - 1 >= 0):
  
        # Appending x-1 to 
        # the current number 
        generateSteppingNos(x * 10 + last - 1, s)
  
    # Appending x to 
    # the current number 
    generateSteppingNos(x * 10 + last, s) 
  
    if (last + 1 <= 9):
      
        # Appending x+1 to 
        # the current number 
        generateSteppingNos(x * 10 + last + 1, s)
  
# Function to print 
# N Stepping numbers 
def NSteppingNumbers(N):
  
    s = set()
  
    for i in range(1, 10): 
        generateSteppingNos(i, s)
  
    count = 1
  
    s.pop()
  
    # Printing N numbers from s 
    for value in s: 
        if (count <= N): 
            print(value, end=', ')
            count = count + 1
        else:
            break
  
# Driver code 
N = 14
  
NSteppingNumbers(N)
  
# This code is contributed by Sanjit_Prasad

输出：

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 21, 22,