最大乘积子数组的 C++ 程序

给定一个包含正整数和负整数的数组，找到最大乘积子数组的乘积。预期时间复杂度为 O(n)，只能使用 O(1) 额外空间。

例子：

Input: arr[] = {6, -3, -10, 0, 2}
Output:   180  // The subarray is {6, -3, -10}

Input: arr[] = {-1, -3, -10, 0, 60}
Output:   60  // The subarray is {60}

Input: arr[] = {-2, -40, 0, -2, -3}
Output:   80  // The subarray is {-2, -40}

天真的解决方案：

这个想法是遍历每个连续的子数组，找到每个子数组的乘积并从这些结果中返回最大乘积。

下面是上述方法的实现。

C++

// C++ program to find Maximum Product Subarray
#include 
using namespace std;
  
/* Returns the product of max product subarray.*/
int maxSubarrayProduct(int arr[], int n)
{
    // Initializing result
    int result = arr[0];
  
    for (int i = 0; i < n; i++) 
    {
        int mul = arr[i];
        // traversing in current subarray
        for (int j = i + 1; j < n; j++) 
        {
            // updating result every time
            // to keep an eye over the maximum product
            result = max(result, mul);
            mul *= arr[j];
        }
        // updating the result for (n-1)th index.
        result = max(result, mul);
    }
    return result;
}
  
// Driver code
int main()
{
    int arr[] = { 1, -2, -3, 0, 7, -8, -2 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << "Maximum Sub array product is "
         << maxSubarrayProduct(arr, n);
    return 0;
}
  
// This code is contributed by yashbeersingh42

输出：

Maximum Sub array product is 112

时间复杂度： O(N ² )
辅助空间： O(1)

高效解决方案：

以下解决方案假定给定的输入数组始终具有正输出。该解决方案适用于上述所有情况。它不适用于 {0, 0, -20, 0}, {0, 0, 0}.. 等数组。可以轻松修改解决方案以处理这种情况。
它类似于最大和连续子数组问题。这里唯一要注意的是，最大乘积也可以通过以前一个元素结尾的最小（负）乘积乘以这个元素来获得。例如，在数组{12, 2, -3, -5, -6, -2}中，当我们在元素-2处时，最大乘积是乘积，最小乘积以-6和-2结尾。

C++

// C++ program to find Maximum Product Subarray
#include 
using namespace std;
  
/* Returns the product 
  of max product subarray.
Assumes that the given 
array always has a subarray
with product more than 1 */
int maxSubarrayProduct(int arr[], int n)
{
    // max positive product 
    // ending at the current position
    int max_ending_here = 1;
  
    // min negative product ending 
    // at the current position
    int min_ending_here = 1;
  
    // Initialize overall max product
    int max_so_far = 0;
    int flag = 0;
    /* Traverse through the array. 
    Following values are
    maintained after the i'th iteration:
    max_ending_here is always 1 or 
    some positive product ending with arr[i]
    min_ending_here is always 1 or 
    some negative product ending with arr[i] */
    for (int i = 0; i < n; i++)
    {
        /* If this element is positive, update
        max_ending_here. Update min_ending_here only if
        min_ending_here is negative */
        if (arr[i] > 0) 
        {
            max_ending_here = max_ending_here * arr[i];
            min_ending_here
                = min(min_ending_here * arr[i], 1);
            flag = 1;
        }
  
        /* If this element is 0, then the maximum product
        cannot end here, make both max_ending_here and
        min_ending_here 0
        Assumption: Output is alway greater than or equal
                    to 1. */
        else if (arr[i] == 0) {
            max_ending_here = 1;
            min_ending_here = 1;
        }
  
        /* If element is negative. This is tricky
         max_ending_here can either be 1 or positive.
         min_ending_here can either be 1 or negative.
         next max_ending_here will always be prev.
         min_ending_here * arr[i] ,next min_ending_here
         will be 1 if prev max_ending_here is 1, otherwise
         next min_ending_here will be prev max_ending_here *
         arr[i] */
  
        else {
            int temp = max_ending_here;
            max_ending_here
                = max(min_ending_here * arr[i], 1);
            min_ending_here = temp * arr[i];
        }
  
        // update max_so_far, if needed
        if (max_so_far < max_ending_here)
            max_so_far = max_ending_here;
    }
    if (flag == 0 && max_so_far == 0)
        return 0;
    return max_so_far;
}
  
// Driver code
int main()
{
    int arr[] = { 1, -2, -3, 0, 7, -8, -2 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << "Maximum Sub array product is "
         << maxSubarrayProduct(arr, n);
    return 0;
}
  
// This is code is contributed by rathbhupendra

输出

Maximum Sub array product is 112

时间复杂度： O(n)
辅助空间： O(1)

有关更多详细信息，请参阅有关最大乘积子阵列的完整文章！