大小为 k 的子数组的最大乘积

给定一个由 n 个正整数和一个整数 k 组成的数组。找到大小为 k 的最大乘积子数组，即在 k <= n 的数组中找到 k 个连续元素的最大乘积。
例子 ：

Input: arr[] = {1, 5, 9, 8, 2, 4,
                 1, 8, 1, 2} 
       k = 6
Output:   4608  
The subarray is {9, 8, 2, 4, 1, 8}

Input: arr[] = {1, 5, 9, 8, 2, 4, 1, 8, 1, 2}
       k = 4
Output:   720  
The subarray is {5, 9, 8, 2}

Input: arr[] = {2, 5, 8, 1, 1, 3};
       k = 3             
Output:   80  
The subarray is {2, 5, 8}

方法1（简单：O（n * k））
一种朴素的方法是一个接一个地考虑所有大小为 k 的子数组。这种方法需要两个循环，因此复杂度为 O(n*k)。
方法2（效率：O（n））
如果我们有可用的前一个子数组的乘积，我们可以使用以下事实在 O(n) 中解决它，即大小为 k 的子数组的乘积可以在 O(1) 时间内计算出来。

curr_product = (prev_product / arr[i-1]) * arr[i + k -1]

prev_product : Product of subarray of size k beginning 
               with arr[i-1]

curr_product : Product of subarray of size k beginning 
               with arr[i]

这样，我们只需一次遍历就可以计算出最大 k 个子数组的乘积。下面是这个想法的 C++ 实现。

C++

// C++ program to find the maximum product of a subarray
// of size k.
#include 
using namespace std;
 
// This function returns maximum product of a subarray
// of size k in given array, arr[0..n-1]. This function
// assumes that k is smaller than or equal to n.
int findMaxProduct(int arr[], int n, int k)
{
    // Initialize the MaxProduct to 1, as all elements
    // in the array are positive
    int MaxProduct = 1;
    for (int i=0; i

---

Java

// Java program to find the maximum product of a subarray
// of size k
import java.io.*;
import java.util.*;
 
class GFG
{
    // Function returns maximum product of a subarray
    // of size k in given array, arr[0..n-1]. This function
    // assumes that k is smaller than or equal to n.
    static int findMaxProduct(int arr[], int n, int k)
    {
        // Initialize the MaxProduct to 1, as all elements
        // in the array are positive
        int MaxProduct = 1;
        for (int i=0; i

---

Python3

# Python 3 program to find the maximum
# product of a subarray of size k.
 
# This function returns maximum product
# of a subarray of size k in given array,
# arr[0..n-1]. This function assumes
# that k is smaller than or equal to n.
def findMaxProduct(arr, n, k) :
   
    # Initialize the MaxProduct to 1,
    # as all elements in the array
    # are positive
    MaxProduct = 1
    for i in range(0, k) :
        MaxProduct = MaxProduct * arr[i]
         
    prev_product = MaxProduct
  
    # Consider every product beginning
    # with arr[i] where i varies from
    # 1 to n-k-1
    for i in range(1, n - k + 1) :
        curr_product = (prev_product // arr[i-1]) * arr[i+k-1]
        MaxProduct = max(MaxProduct, curr_product)
        prev_product = curr_product
     
     
    # Return the maximum product found
    return MaxProduct
     
# Driver code
arr1 = [1, 5, 9, 8, 2, 4, 1, 8, 1, 2]
k = 6
n = len(arr1)
print (findMaxProduct(arr1, n, k) )
k = 4
print (findMaxProduct(arr1, n, k))
 
arr2 = [2, 5, 8, 1, 1, 3]
k = 3
n = len(arr2)
 
print(findMaxProduct(arr2, n, k))
 
# This code is contributed by Nikita Tiwari.

C#

// C# program to find the maximum
// product of a subarray of size k
using System;
 
class GFG
{
    // Function returns maximum
    // product of a subarray of
    // size k in given array,
    // arr[0..n-1]. This function
    // assumes that k is smaller
    // than or equal to n.
    static int findMaxProduct(int []arr,
                              int n, int k)
    {
        // Initialize the MaxProduct
        // to 1, as all elements
        // in the array are positive
        int MaxProduct = 1;
        for (int i = 0; i < k; i++)
            MaxProduct *= arr[i];
 
        int prev_product = MaxProduct;
 
        // Consider every product beginning
        // with arr[i] where i varies from
        // 1 to n-k-1
        for (int i = 1; i <= n - k; i++)
        {
            int curr_product = (prev_product /
                                 arr[i - 1]) *
                                 arr[i + k - 1];
            MaxProduct = Math.Max(MaxProduct,
                                  curr_product);
            prev_product = curr_product;
        }
 
        // Return the maximum
        // product found
        return MaxProduct;
    }
     
    // Driver Code
    public static void Main ()
    {
        int []arr1 = {1, 5, 9, 8, 2,
                      4, 1, 8, 1, 2};
        int k = 6;
        int n = arr1.Length;
        Console.WriteLine(findMaxProduct(arr1, n, k));
 
        k = 4;
        Console.WriteLine(findMaxProduct(arr1, n, k));
 
        int []arr2 = {2, 5, 8, 1, 1, 3};
        k = 3;
        n = arr2.Length;
        Console.WriteLine(findMaxProduct(arr2, n, k));
    }
}
 
// This code is contributed by anuj_67.

Javascript

 
    // JavaScript program to find the maximum
    // product of a subarray of size k
     
    // Function returns maximum
    // product of a subarray of
    // size k in given array,
    // arr[0..n-1]. This function
    // assumes that k is smaller
    // than or equal to n.
    function findMaxProduct(arr, n, k)
    {
        // Initialize the MaxProduct
        // to 1, as all elements
        // in the array are positive
        let MaxProduct = 1;
        for (let i = 0; i < k; i++)
            MaxProduct *= arr[i];
   
        let prev_product = MaxProduct;
   
        // Consider every product beginning
        // with arr[i] where i varies from
        // 1 to n-k-1
        for (let i = 1; i <= n - k; i++)
        {
            let curr_product =
            (prev_product / arr[i - 1]) * arr[i + k - 1];
            MaxProduct = Math.max(MaxProduct, curr_product);
            prev_product = curr_product;
        }
   
        // Return the maximum
        // product found
        return MaxProduct;
    }
     
    let arr1 = [1, 5, 9, 8, 2, 4, 1, 8, 1, 2];
    let k = 6;
    let n = arr1.length;
    document.write(findMaxProduct(arr1, n, k) + "</br>");
 
    k = 4;
    document.write(findMaxProduct(arr1, n, k) + "</br>");
 
    let arr2 = [2, 5, 8, 1, 1, 3];
    k = 3;
    n = arr2.length;
    document.write(findMaxProduct(arr2, n, k) + "</br>");
     

输出：

4608
720
80