在小于或等于N的数字中查找数字的最大乘积

给定整数N> 0 ，任务是在小于或等于N的数字中找到数字的最大乘积。
例子：

Input: N = 390 
Output: 216 
Maximum possible product is given by the number 389 
3 * 8 * 9 = 216
Input: N = 432 
Output: 243 
 

方法：也可以使用本文中描述的方法解决此问题，将下限设为1 ，将上限设为N。解决此问题的另一种方法是使用递归。递归的条件如下：

• 如果N = 0，则返回1 。
• 如果N <10，则返回N。
• 否则，返回max(maxProd(N / 10)*(N％10)，maxProd((N / 10)– 1)* 9

在递归的每个步骤中，采用最后一位或9位来最大化位的乘积。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function that returns the maximum product of
// digits among numbers less than or equal to N
int maxProd(int N)
{
    if (N == 0)
        return 1;
    if (N < 10)
        return N;
    return max(maxProd(N / 10) * (N % 10),
               maxProd(N / 10 - 1) * 9);
}
 
// Driver code
int main()
{
    int N = 390;
    cout << maxProd(N);
 
    return 0;
}

Java

// Java implementation of the approach
import java.io.*;
 
class GFG
{
     
// Function that returns the maximum product of
// digits among numbers less than or equal to N
static int maxProd(int N)
{
    if (N == 0)
        return 1;
    if (N < 10)
        return N;
    return Math.max(maxProd(N / 10) * (N % 10),
            maxProd(N / 10 - 1) * 9);
}
 
// Driver code
public static void main (String[] args)
{
    int N = 390;
    System.out.println (maxProd(N));
}
}
 
// This code is contributed by ajit.

Python3

# Python3 implementation of the approach
 
# Function that returns the maximum product of
# digits among numbers less than or equal to N
def maxProd(N):
 
    if (N == 0):
        return 1
    if (N < 10):
        return N
    return max(maxProd(N // 10) * (N % 10),
               maxProd(N // 10 - 1) * 9)
 
# Driver code
N = 390
print(maxProd(N))
 
# This code is contributed by mohit kumar

C#

// C# implementation of the approach
using System;
 
class GFG
{
         
// Function that returns the maximum product of
// digits among numbers less than or equal to N
static int maxProd(int N)
{
    if (N == 0)
        return 1;
    if (N < 10)
        return N;
    return Math.Max(maxProd(N / 10) * (N % 10),
            maxProd(N / 10 - 1) * 9);
}
 
// Driver code
static public void Main ()
{
    int N = 390;
    Console.WriteLine(maxProd(N));
}
}
 
// This code is contributed by Tushil..

Javascript

 
// Javascript implementation of the approach
 
// Function that returns the maximum product of
// digits among numbers less than or equal to N
function maxProd(N)
{
    if (N == 0)
        return 1;
    if (N < 10)
        return N;
    return Math.max(maxProd(parseInt(N / 10)) * (N % 10),
            maxProd(parseInt(N / 10) - 1) * 9);
}
 
// Driver code
let N = 390;
document.write(maxProd(N));
 
// This code is contributed
// by bobby
 

输出：

216