原文:
给定一个自然数 n ，任务是打印第 n 个。
如果所有相邻数字的绝对差值都为 1，则该数字称为步进数。以下系列为步进自然数列表: 1、2、3、4、5、6、7、8、9、10、11、12、21、22、23、32、…。
例:
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 数据结构来解决。首先，准备一个空队列，按照的顺序将 1，2，…，9 入队。 然后为了生成第 n 个步进号，必须执行以下操作 n 次:
- 从队列中执行出列。设 x 为出列元素。
- 如果 x mod 10 不等于 0，则入队 10x (x mod 10)–1
- enqueue 10x (x mod 10)。
- 如果 x mod 10 不等于 9，那么入队 10x (x mod 10) 1。
第 n 个操作中的出列号是第 n 个步进号。 以下是上述方法的实施:
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 语言(一种计算机语言，尤用于创建网站)
// 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
python 3
# 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
java 描述语言
output:
32
时间复杂度: o(n)
辅助空间:o(n)t4】
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