给定自然数N ，任务是打印第N个步进数或自传数。
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：使用队列数据结构可以解决此问题。首先，准备一个空队列，然后按此顺序将1、2，…，9入队。
然后，为了生成第N个步进编号，必须将以下操作执行N次：
- 从队列中执行出队。令x为出队元素。
- 如果x mod 10不等于0，则入队10x +(x mod 10)– 1
- 入队10x +(x mod 10)。
- 如果x mod 10不等于9，则入队10x +(x mod 10)+ 1。
在第N个操作中出队的编号是第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_GangwarC#
// 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 输出：
32
时间复杂度： O(N)