打印第 N 个步进或自传编号

给定一个自然数N ，任务是打印第 N 个步进或自传数字。

A number is called stepping number if all adjacent digits have an absolute difference of 1. The following series is a list of Stepping natural numbers: 
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 21, 22, 23, 32, ….

例子：

Input: N = 16 
Output: 32
Explanation:
16th Stepping number is 32.

Input: N = 14 
Output: 22
Explanation:
14th Stepping number is 22.

方法：这个问题可以使用Queue数据结构来解决。首先准备一个空队列，按这个顺序入队1、2、……、9 。
然后为了生成第 N 个 Stepping 数，必须执行以下操作 N 次：

• 从队列中执行出队。令 x 为出列元素。
• 如果 x mod 10 不等于 0，则 Enqueue 10x + (x mod 10) – 1
• 入队 10x + (x mod 10)。
• 如果 x mod 10 不等于 9，则 Enqueue 10x + (x mod 10) + 1。

第 N 次操作中出队的编号为第 N 个步进编号。
下面是上述方法的实现：

C++

// C++ implementation to find
// N’th stepping natural Number
#include 
using namespace std;
 
// Function to find the
// Nth stepping natural number
int NthSmallest(int K)
{
 
    // Declare the queue
    queue Q;
 
    int x;
 
    // Enqueue 1, 2, ..., 9 in this order
    for (int i = 1; i < 10; i++)
        Q.push(i);
 
    // Perform K operation on queue
    for (int i = 1; i <= K; i++) {
 
        // Get the ith Stepping number
        x = Q.front();
 
        // Perform Dequeue from the Queue
        Q.pop();
 
        // If x mod 10 is not equal to 0
        if (x % 10 != 0) {
 
            // then Enqueue 10x + (x mod 10) - 1
            Q.push(x * 10 + x % 10 - 1);
        }
 
        // Enqueue 10x + (x mod 10)
        Q.push(x * 10 + x % 10);
 
        // If x mod 10 is not equal to 9
        if (x % 10 != 9) {
 
            // then Enqueue 10x + (x mod 10) + 1
            Q.push(x * 10 + x % 10 + 1);
        }
    }
 
    // Return the dequeued number of the K-th
    // operation as the Nth stepping number
    return x;
}
 
// Driver Code
int main()
{
 
    // initialise K
    int N = 16;
 
    cout << NthSmallest(N) << "\n";
 
    return 0;
}

Java

// Java implementation to find
// N'th stepping natural Number
import java.util.*;
 
class GFG{
  
// Function to find the
// Nth stepping natural number
static int NthSmallest(int K)
{
  
    // Declare the queue
    Queue Q = new LinkedList<>();
  
    int x = 0;
  
    // Enqueue 1, 2, ..., 9 in this order
    for (int i = 1; i < 10; i++)
        Q.add(i);
  
    // Perform K operation on queue
    for (int i = 1; i <= K; i++) {
  
        // Get the ith Stepping number
        x = Q.peek();
  
        // Perform Dequeue from the Queue
        Q.remove();
  
        // If x mod 10 is not equal to 0
        if (x % 10 != 0) {
  
            // then Enqueue 10x + (x mod 10) - 1
            Q.add(x * 10 + x % 10 - 1);
        }
  
        // Enqueue 10x + (x mod 10)
        Q.add(x * 10 + x % 10);
  
        // If x mod 10 is not equal to 9
        if (x % 10 != 9) {
  
            // then Enqueue 10x + (x mod 10) + 1
            Q.add(x * 10 + x % 10 + 1);
        }
    }
  
    // Return the dequeued number of the K-th
    // operation as the Nth stepping number
    return x;
}
  
// Driver Code
public static void main(String[] args)
{
  
    // initialise K
    int N = 16;
  
    System.out.print(NthSmallest(N));
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation to find
# N’th stepping natural Number
 
# Function to find the
# Nth stepping natural number
def NthSmallest(K):
    # Declare the queue
    Q = []
 
    # Enqueue 1, 2, ..., 9 in this order
    for i in range(1,10):
        Q.append(i)
 
    # Perform K operation on queue
    for i in range(1,K+1):
        # Get the ith Stepping number
        x = Q[0]
 
        # Perform Dequeue from the Queue
        Q.remove(Q[0])
 
        # If x mod 10 is not equal to 0
        if (x % 10 != 0):
            # then Enqueue 10x + (x mod 10) - 1
            Q.append(x * 10 + x % 10 - 1)
 
        # Enqueue 10x + (x mod 10)
        Q.append(x * 10 + x % 10)
 
        # If x mod 10 is not equal to 9
        if (x % 10 != 9):
            # then Enqueue 10x + (x mod 10) + 1
            Q.append(x * 10 + x % 10 + 1)
 
    # Return the dequeued number of the K-th
    # operation as the Nth stepping number
    return x
 
# Driver Code
if __name__ == '__main__':
    # initialise K
    N = 16
 
    print(NthSmallest(N))
 
# This code is contributed by Surendra_Gangwar

C#

// C# implementation to find
// N'th stepping natural Number
using System;
using System.Collections.Generic;
 
class GFG{
   
// Function to find the
// Nth stepping natural number
static int NthSmallest(int K)
{
   
    // Declare the queue
    List Q = new List();
   
    int x = 0;
   
    // Enqueue 1, 2, ..., 9 in this order
    for (int i = 1; i < 10; i++)
        Q.Add(i);
   
    // Perform K operation on queue
    for (int i = 1; i <= K; i++) {
   
        // Get the ith Stepping number
        x = Q[0];
   
        // Perform Dequeue from the Queue
        Q.RemoveAt(0);
   
        // If x mod 10 is not equal to 0
        if (x % 10 != 0) {
   
            // then Enqueue 10x + (x mod 10) - 1
            Q.Add(x * 10 + x % 10 - 1);
        }
   
        // Enqueue 10x + (x mod 10)
        Q.Add(x * 10 + x % 10);
   
        // If x mod 10 is not equal to 9
        if (x % 10 != 9) {
   
            // then Enqueue 10x + (x mod 10) + 1
            Q.Add(x * 10 + x % 10 + 1);
        }
    }
   
    // Return the dequeued number of the K-th
    // operation as the Nth stepping number
    return x;
}
   
// Driver Code
public static void Main(String[] args)
{
   
    // initialise K
    int N = 16;
   
    Console.Write(NthSmallest(N));
}
}
 
// This code is contributed by sapnasingh4991

Javascript

 
// Javascript implementation to find
// N’th stepping natural Number
 
// Function to find the
// Nth stepping natural number
function NthSmallest(K)
{
 
    // Declare the queue
    var Q = [];
 
    var x;
 
    // Enqueue 1, 2, ..., 9 in this order
    for (var i = 1; i < 10; i++)
        Q.push(i);
 
    // Perform K operation on queue
    for (var i = 1; i <= K; i++) {
 
        // Get the ith Stepping number
        x = Q[0];
 
        // Perform Dequeue from the Queue
        Q.shift();
 
        // If x mod 10 is not equal to 0
        if (x % 10 != 0) {
 
            // then Enqueue 10x + (x mod 10) - 1
            Q.push(x * 10 + x % 10 - 1);
        }
 
        // Enqueue 10x + (x mod 10)
        Q.push(x * 10 + x % 10);
 
        // If x mod 10 is not equal to 9
        if (x % 10 != 9) {
 
            // then Enqueue 10x + (x mod 10) + 1
            Q.push(x * 10 + x % 10 + 1);
        }
    }
 
    // Return the dequeued number of the K-th
    // operation as the Nth stepping number
    return x;
}
 
// Driver Code
// initialise K
var N = 16;
document.write( NthSmallest(N));
 
// This code is contributed by famously.

输出：

32

时间复杂度： O(N)