第一个数组中的最大值与第二个数组中的最小值的乘积 · GeeksForGeeks 中文系列教程

# 第一个数组中的最大值与第二个数组中的最小值的乘积 > 原文： [https://www.geeksforgeeks.org/product-maximum-first-array-minimum-second/](https://www.geeksforgeeks.org/product-maximum-first-array-minimum-second/) 给定两个数组，任务是计算第一个数组的 max 元素和第二个数组的 min 元素的乘积参考文献：在 Adobe 中提问（来源： [Careercup](https://www.careercup.com/question?id=5703822780399616) ） **示例**： ``` Input : arr1[] = {5, 7, 9, 3, 6, 2}, arr2[] = {1, 2, 6, -1, 0, 9} Output : max element in first array is 9 and min element in second array is -1\. The product of these two is -9. Input : arr1[] = {1, 4, 2, 3, 10, 2}, arr2[] = {4, 2, 6, 5, 2, 9} Output : max element in first array is 10 and min element in second array is 2\. The product of these two is 20. ``` **方法 1：简单方法**我们首先对两个数组进行排序。然后，我们可以轻松地在第一个数组中找到 max，在第二个数组中找到 min。最后，我们返回最小值和最大值的乘积。 ## C++ ```cpp // C++ program to calculate the // product of max element of // first array and min element // of second array #include <bits/stdc++.h> using namespace std; // Function to calculate // the product int minMaxProduct(int arr1[], int arr2[], int n1, int n2) { // Sort the arrays to find // the maximum and minimum // elements in given arrays sort(arr1, arr1 + n1); sort(arr2, arr2 + n2); // Return product of // maximum and minimum. return arr1[n1 - 1] * arr2[0]; } // Driven code int main() { int arr1[] = { 10, 2, 3, 6, 4, 1 }; int arr2[] = { 5, 1, 4, 2, 6, 9 }; int n1 = sizeof(arr1) / sizeof(arr1[0]); int n2 = sizeof(arr1) / sizeof(arr1[0]); cout << minMaxProductt(arr1, arr2, n1, n2); return 0; } ``` ## Java ```java // Java program to find the // to calculate the product // of max element of first // array and min element of // second array import java.util.*; import java.lang.*; class GfG { // Function to calculate // the product public static int minMaxProduct(int arr1[], int arr2[], int n1, int n2) { // Sort the arrays to find the // maximum and minimum elements // in given arrays Arrays.sort(arr1); Arrays.sort(arr2); // Return product of maximum // and minimum. return arr1[n1 - 1] * arr2[0]; } // Driver Code public static void main(String argc[]) { int [] arr1= new int []{ 10, 2, 3, 6, 4, 1 }; int [] arr2 = new int []{ 5, 1, 4, 2, 6, 9 }; int n1 = 6; int n2 = 6; System.out.println(minMaxProduct(arr1, arr2, n1, n2)); } } /*This code is contributed by Sagar Shukla.*/ ``` ## Python ``` # A Python program to find the to # calculate the product of max # element of first array and min # element of second array # Function to calculate the product def minmaxProduct(arr1, arr2, n1, n2): # Sort the arrays to find the # maximum and minimum elements # in given arrays arr1.sort() arr2.sort() # Return product of maximum # and minimum. return arr1[n1 - 1] * arr2[0] # Driver Program arr1 = [10, 2, 3, 6, 4, 1] arr2 = [5, 1, 4, 2, 6, 9] n1 = len(arr1) n2 = len(arr2) print(minmaxProduct(arr1, arr2, n1, n2)) # This code is contributed by Shrikant13\. ``` ## C# ```cs // C# program to find the to // calculate the product of // max element of first array // and min element of second array using System; class GfG { // Function to calculate the product public static int minMaxProduct(int []arr1, int []arr2, int n1, int n2) { // Sort the arrays to find the // maximum and minimum elements // in given arrays Array.Sort(arr1); Array.Sort(arr2); // Return product of maximum // and minimum. return arr1[n1 - 1] * arr2[0]; } // Driver Code public static void Main() { int [] arr1= new int []{ 10, 2, 3, 6, 4, 1 }; int [] arr2 = new int []{ 5, 1, 4, 2, 6, 9 }; int n1 = 6; int n2 = 6; Console.WriteLine(minMaxProduct(arr1, arr2, n1, n2)); } } /*This code is contributed by vt_m.*/ ``` ## PHP ```php <?php // PHP program to find the to // calculate the product of max // element of first array and // min element of second array // Function to calculate the product function minMaxProduct( $arr1, $arr2, $n1, $n2) { // Sort the arrays to find // the maximum and minimum // elements in given arrays sort($arr1); sort($arr2); // Return product of // maximum and minimum. return $arr1[$n1 - 1] * $arr2[0]; } // Driver code $arr1 = array( 10, 2, 3, 6, 4, 1 ); $arr2 = array( 5, 1, 4, 2, 6, 9 ); $n1 = count($arr1); $n2 = count($arr2); echo minMaxProduct($arr1, $arr2, $n1, $n2); // This code is contributed by anuj_67\. ?> ``` **输出**： ``` 10 ``` **时间复杂度**：`O(N log N)` **空间复杂度**：`O(1)` **方法 2：高效方法**在这种方法中，我们只需遍历整个数组，然后在第一个数组中找到 max，在第二个数组中找到 min，就可以轻松获得 min 和 max 的乘积。 ## C++ ``` // C++ program to find the to // calculate the product of // max element of first array // and min element of second array #include <bits/stdc++.h> using namespace std; // Function to calculate the product int minMaxProduct(int arr1[], int arr2[], int n1, int n2) { // Initialize max of first array int max = arr1[0]; // initialize min of second array int min = arr2[0]; int i; for (i = 1; i < n1 && i < n2; ++i) { // To find the maximum // element in first array if (arr1[i] > max) max = arr1[i]; // To find the minimum // element in second array if (arr2[i] < min) min = arr2[i]; } // Process remaining elements while (i < n1) { if (arr1[i] > max) max = arr1[i]; i++; } while (i < n2) { if (arr2[i] < min) min = arr2[i]; i++; } return max * min; } // Driven code int main() { int arr1[] = { 10, 2, 3, 6, 4, 1 }; int arr2[] = { 5, 1, 4, 2, 6, 9 }; int n1 = sizeof(arr1) / sizeof(arr1[0]); int n2 = sizeof(arr1) / sizeof(arr1[0]); cout << minMaxProduct(arr1, arr2, n1, n2) << endl; return 0; } ```