计算所有排列大于该数的自然数

有一些自然数的所有排列都大于或等于该数，例如。 123，其所有排列（123、231、321）都大于或等于123。
给定一个自然数n ，任务是计算从 1 到 n 的所有这些数字。

例子：

Input : n = 15.
Output : 14
Explanation:
1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 
13, 14, 15 are the numbers whose all 
permutation is greater than the number
itself. So, output 14.

Input : n = 100.
Output : 54

一个简单的解决方案是运行一个从 1 到 n 的循环，并为每个数字检查其数字是否处于非递减顺序。

一个有效的解决方案基于以下观察。
观察 1：从 1 到 9，所有数字都有这个性质。因此，对于 n <= 9，输出 n。
观察 2：所有排列大于或等于该数的数的所有数字都按升序排列。
这个想法是将所有数字从 1 推到 9。现在，弹出顶部元素，说topel并尝试制作数字按升序排列的数字，第一个数字是topel 。要制作这样的数字，第二个数字可以从topel%10到 9。如果这个数字小于n ，则增加计数并将数字压入堆栈，否则忽略。

下面是这种方法的实现：

C++

// C++ program to count the number less than N,
// whose all permutation is greater
// than or equal to the number.
#include 
using namespace std;
 
// Return the count of the number having all
// permutation greater than or equal to the number.
int countNumber(int n)
{
    int result = 0;
 
    // Pushing 1 to 9 because all number from 1
    // to 9 have this property.
    stack s;
    for (int i = 1; i <= 9; i++)
    {
         
        if (i <= n)
        {
            s.push(i);
            result++;
        }
 
        // take a number from stack and add
        // a digit smaller than or equal to last digit
        // of it.
        while (!s.empty())
        {
            int tp = s.top();
            s.pop();
            for (int j = tp % 10; j <= 9; j++)
            {
                int x = tp * 10 + j;
                if (x <= n)
                {
                    s.push(x);
                    result++;
                }
            }
        }
    }
 
    return result;
}
 
// Driven Code
int main()
{
    int n = 15;
    cout << countNumber(n) << endl;
    return 0;
}

Java

// Java program to count the number less than N,
// whose all permutation is greater
// than or equal to the number.
import java.util.Stack;
 
 
class GFG
{
    // Return the count of the number having all
    // permutation greater than or equal to the number.
 
    static int countNumber(int n)
    {
        int result = 0;
 
        // Pushing 1 to 9 because all number from 1
        // to 9 have this property.
        Stack s = new Stack<>();
        for (int i = 1; i <= 9; i++)
        {
 
            if (i <= n)
            {
                s.push(i);
                result++;
            }
 
            // take a number from stack and add
            // a digit smaller than or equal to last digit
            // of it.
            while (!s.empty())
            {
                int tp = s.pop();
                
                for (int j = tp % 10; j <= 9; j++)
                {
                    int x = tp * 10 + j;
                    if (x <= n) {
                        s.push(x);
                        result++;
                    }
                }
            }
        }
 
        return result;
    }
 
    // Driven Code
    public static void main(String[] args)
    {
        int n = 15;
        System.out.println(countNumber(n));
    }
}
 
// this code contributed by Rajput-Ji

Python3

# Python3 program to count the number less
# than N, whose all permutation is greater
# than or equal to the number.
 
# Return the count of the number having
# all permutation greater than or equal
# to the number.
 
 
def countNumber(n):
    result = 0
 
    # Pushing 1 to 9 because all number
    # from 1 to 9 have this property.
    s = []
    for i in range(1, 10):
         
        if (i <= n):
            s.append(i)
            result += 1
 
        # take a number from stack and add
        # a digit smaller than or equal to last digit
        # of it.
        while len(s) != 0:
            tp = s[-1]
            s.pop()
            for j in range(tp % 10, 10):
                x = tp * 10 + j
                if (x <= n):
                    s.append(x)
                    result += 1
 
    return result
 
 
# Driver Code
if __name__ == '__main__':
 
    n = 15
    print(countNumber(n))
 
# This code is contributed by PranchalK

C#

// C# program to count the number less than N,
// whose all permutation is greater than
// or equal to the number.
using System;
using System.Collections.Generic;
 
class GFG {
 
    // Return the count of the number
    // having all permutation greater than
    // or equal to the number.
    static int countNumber(int n)
    {
        int result = 0;
 
        // Pushing 1 to 9 because all number from 1
        // to 9 have this property.
        Stack s = new Stack();
        for (int i = 1; i <= 9; i++)
        {
             
            if (i <= n)
            {
                s.Push(i);
                result++;
            }
 
            // take a number from stack and add
            // a digit smaller than or equal to last digit
            // of it.
            while (s.Count != 0)
            {
                int tp = s.Peek();
                s.Pop();
                for (int j = tp % 10; j <= 9; j++)
                {
                    int x = tp * 10 + j;
                    if (x <= n) {
                        s.Push(x);
                        result++;
                    }
                }
            }
        }
 
        return result;
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
        int n = 15;
        Console.WriteLine(countNumber(n));
    }
}
 
// This code is contributed by Rajput-Ji

Javascript

 
        // JavaScript program for the above approach
 
        // Return the count of the number having all
        // permutation greater than or equal to the number.
        function countNumber(n)
        {
            let result = 0;
 
            // Pushing 1 to 9 because all number from 1
            // to 9 have this property.
            let s = [];
            for (let i = 1; i <= 9; i++)
            {
 
                if (i <= n)
                {
                    s.push(i);
                    result++;
                }
 
                // take a number from stack and add
                // a digit smaller than or equal to last digit
                // of it.
                while (s.length != 0)
                {
                    let tp = s[s.length - 1];
                    s.pop();
                    for (let j = tp % 10; j <= 9; j++)
                    {
                        let x = tp * 10 + j;
                        if (x <= n)
                        {
                            s.push(x);
                            result++;
                        }
                    }
                }
            }
 
            return result;
        }
 
        // Driven Code
 
        let n = 15;
        document.write(countNumber(n));
 
// This code is contributed by Potta Lokesh
    

输出

14

时间复杂度： O(x)，其中 x 是输出中打印的元素数。