矩阵中 4 个相邻元素的最大积 · GeeksForGeeks 中文系列教程

# 矩阵中 4 个相邻元素的最大积 > 原文： [https://www.geeksforgeeks.org/maximum-product-of-4-adjacent-elements-in-matrix/](https://www.geeksforgeeks.org/maximum-product-of-4-adjacent-elements-in-matrix/) 给定一个正方形矩阵，找到矩阵的四个相邻元素的最大积。矩阵的相邻元素可以是上，下，左，右，对角线或反对角线。四个或更多数字应彼此相邻。 **注意**： n 应该大于或等于 4，即> = 4 **示例**： ``` Input : n = 4 {{6, 2, 3 4}, {5, 4, 3, 1}, {7, 4, 5, 6}, {8, 3, 1, 0}} Output : 1680 Explanation: Multiplication of 6 5 7 8 produces maximum result and all element are adjacent to each other in one direction Input : n = 5 {{1, 2, 3, 4, 5}, {6, 7, 8, 9, 1}, {2, 3, 4, 5, 6}, {7, 8, 9, 1, 0}, {9, 6, 4, 2, 3}} Output: 3024 Explanation: Multiplication of 6 7 8 9 produces maximum result and all elements are adjacent to each other in one direction. ``` **提问者：Tolexo** **方法**： 1.将每行中彼此相邻的 4 个元素分组，并计算其最大结果。 2.在每列中彼此相邻的第 4 组元素，并计算其最大结果。 3.对角线中彼此相邻的第 4 组元素，并计算其最大结果。 4.以对角线方向彼此相邻的第 4 组元素，并计算其最大结果。 5.比较所有计算出的最大结果。下面是上述方法的实现： ## C++ ```cpp // C++ program to find out the maximum product // in the matrix which four elements are // adjacent to each other in one direction #include <bits/stdc++.h> using namespace std; const int n = 5; // function to find max product int FindMaxProduct(int arr[][n], int n) { int max = 0, result; // iterate the rows. for (int i = 0; i < n; i++) { // iterate the columns. for (int j = 0; j < n; j++) { // check the maximum product // in horizontal row. if ((j - 3) >= 0) { result = arr[i][j] * arr[i][j - 1] * arr[i][j - 2] * arr[i][j - 3]; if (max < result) max = result; } // check the maximum product // in vertical row. if ((i - 3) >= 0) { result = arr[i][j] * arr[i - 1][j] * arr[i - 2][j] * arr[i - 3][j]; if (max < result) max = result; } // check the maximum product in // diagonal (going through down - right) if ((i - 3) >= 0 && (j - 3) >= 0) { result = arr[i][j] * arr[i - 1][j - 1] * arr[i - 2][j - 2] * arr[i - 3][j - 3]; if (max < result) max = result; } // check the maximum product in // diagonal (going through up - right) if ((i - 3) >= 0 && (j - 1) <= 0) { result = arr[i][j] * arr[i - 1][j + 1] * arr[i - 2][j + 2] * arr[i - 3][j + 3]; if (max < result) max = result; } } } return max; } // driver code int main() { /* int arr[][4] = {{6, 2, 3, 4}, {5, 4, 3, 1}, {7, 4, 5, 6}, {8, 3, 1, 0}};*/ /* int arr[][5] = {{1, 2, 1, 3, 4}, {5, 6, 3, 9, 2}, {7, 8, 8, 1, 2}, {1, 0, 7, 9, 3}, {3, 0, 8, 4, 9}};*/ int arr[][5] = {{1, 2, 3, 4, 5}, {6, 7, 8, 9, 1}, {2, 3, 4, 5, 6}, {7, 8, 9, 1, 0}, {9, 6, 4, 2, 3}}; cout << FindMaxProduct(arr, n); return 0; } ``` ## Java ```java // Java program to find out the // maximum product in the matrix // which four elements are adjacent // to each other in one direction class GFG { static final int n = 5; // function to find max product static int FindMaxProduct(int arr[][], int n) { int max = 0, result; // iterate the rows. for (int i = 0; i < n; i++) { // iterate the columns. for (int j = 0; j < n; j++) { // check the maximum product // in horizontal row. if ((j - 3) >= 0) { result = arr[i][j] * arr[i][j - 1] * arr[i][j - 2] * arr[i][j - 3]; if (max < result) max = result; } // check the maximum product // in vertical row. if ((i - 3) >= 0) { result = arr[i][j] * arr[i - 1][j] * arr[i - 2][j] * arr[i - 3][j]; if (max < result) max = result; } // check the maximum product in // diagonal (going through down - right) if ((i - 3) >= 0 && (j - 3) >= 0) { result = arr[i][j] * arr[i - 1][j - 1] * arr[i - 2][j - 2] * arr[i - 3][j - 3]; if (max < result) max = result; } // check the maximum product in // diagonal (going through up - right) if ((i - 3) >= 0 && (j - 1) <= 0) { result = arr[i][j] * arr[i - 1][j + 1] * arr[i - 2][j + 2] * arr[i - 3][j + 3]; if (max < result) max = result; } } } return max; } // Driver code public static void main(String[] args) { /* int arr[][4] = {{6, 2, 3, 4}, {5, 4, 3, 1}, {7, 4, 5, 6}, {8, 3, 1, 0}};*/ /* int arr[][5] = {{1, 2, 1, 3, 4}, {5, 6, 3, 9, 2}, {7, 8, 8, 1, 2}, {1, 0, 7, 9, 3}, {3, 0, 8, 4, 9}};*/ int arr[][] = {{1, 2, 3, 4, 5}, {6, 7, 8, 9, 1}, {2, 3, 4, 5, 6}, {7, 8, 9, 1, 0}, {9, 6, 4, 2, 3}}; System.out.print(FindMaxProduct(arr, n)); } } // This code is contributed by Anant Agarwal. ``` ## Python 3 ``` # Python 3 program to find out the maximum # product in the matrix which four elements # are adjacent to each other in one direction n = 5 # function to find max product def FindMaxProduct(arr, n): max = 0 # iterate the rows. for i in range(n): # iterate the columns. for j in range( n): # check the maximum product # in horizontal row. if ((j - 3) >= 0): result = (arr[i][j] * arr[i][j - 1] * arr[i][j - 2] * arr[i][j - 3]) if (max < result): max = result # check the maximum product # in vertical row. if ((i - 3) >= 0) : result = (arr[i][j] * arr[i - 1][j] * arr[i - 2][j] * arr[i - 3][j]) if (max < result): max = result # check the maximum product in # diagonal going through down - right if ((i - 3) >= 0 and (j - 3) >= 0): result = (arr[i][j] * arr[i - 1][j - 1] * arr[i - 2][j - 2] * arr[i - 3][j - 3]) if (max < result): max = result # check the maximum product in # diagonal going through up - right if ((i - 3) >= 0 and (j - 1) <= 0): result = (arr[i][j] * arr[i - 1][j + 1] * arr[i - 2][j + 2] * arr[i - 3][j + 3]) if (max < result): max = result return max # Driver code if __name__ == "__main__": # int arr[][4] = {{6, 2, 3, 4}, # {5, 4, 3, 1}, # {7, 4, 5, 6}, # {8, 3, 1, 0}}; # int arr[][5] = {{1, 2, 1, 3, 4}, # {5, 6, 3, 9, 2}, # {7, 8, 8, 1, 2}, # {1, 0, 7, 9, 3}, # {3, 0, 8, 4, 9}}; arr = [[1, 2, 3, 4, 5], [6, 7, 8, 9, 1], [2, 3, 4, 5, 6], [7, 8, 9, 1, 0], [9, 6, 4, 2, 3]] print(FindMaxProduct(arr, n)) # This code is contributed by ita_c ``` ## C# ```cs // C# program to find out the // maximum product in the matrix // which four elements are adjacent // to each other in one direction using System; public class GFG { static int n = 5; // Function to find max product static int FindMaxProduct(int[,] arr, int n) { int max = 0, result; // iterate the rows for (int i = 0; i < n; i++) { // iterate the columns for (int j = 0; j < n; j++) { // check the maximum product // in horizontal row. if ((j - 3) >= 0) { result = arr[i, j] * arr[i, j - 1] * arr[i, j - 2] * arr[i, j - 3]; if (max < result) max = result; } // check the maximum product // in vertical row. if ((i - 3) >= 0) { result = arr[i, j] * arr[i - 1, j] * arr[i - 2, j] * arr[i - 3, j]; if (max < result) max = result; } // check the maximum product in // diagonal going through down - right if ((i - 3) >= 0 && (j - 3) >= 0) { result = arr[i, j] * arr[i - 1, j - 1] * arr[i - 2, j - 2] * arr[i - 3, j - 3]; if (max < result) max = result; } // check the maximum product in // diagonal going through up - right if ((i - 3 ) >= 0 && (j - 1) <= 0) { result = arr[i, j] * arr[i - 1, j + 1] * arr[i - 2, j + 2] * arr[i - 3, j + 3]; if (max < result) max = result; } } } return max; } // Driver Code static public void Main () { int[,]arr = {{1, 2, 3, 4, 5}, {6, 7, 8, 9, 1}, {2, 3, 4, 5, 6}, {7, 8, 9, 1, 0}, {9, 6, 4, 2, 3}}; Console.Write(FindMaxProduct(arr, n)); } } // This code is contributed by Shrikant13 ``` ## PHP ```php <?php // PHP program to find out the maximum product // in the matrix which four elements are // adjacent to each other in one direction $n = 5; // function to find max product function FindMaxProduct( $arr, $n) { $max = 0; $result; // iterate the rows. for ( $i = 0; $i < $n; $i++) { // iterate the columns. for ( $j = 0; $j < $n; $j++) { // check the maximum product // in horizontal row. if (($j - 3) >= 0) { $result = $arr[$i][$j] * $arr[$i][$j - 1] * $arr[$i][$j - 2] * $arr[$i][$j - 3]; if ($max < $result) $max = $result; } // check the maximum product // in vertical row. if (($i - 3) >= 0) { $result = $arr[$i][$j] * $arr[$i - 1][$j] * $arr[$i - 2][$j] * $arr[$i - 3][$j]; if ($max < $result) $max = $result; } // check the maximum product in // diagonal going through down - right if (($i - 3) >= 0 and ($j - 3) >= 0) { $result = $arr[$i][$j] * $arr[$i - 1][$j - 1] * $arr[$i - 2][$j - 2] * $arr[$i - 3][$j - 3]; if ($max < $result) $max = $result; } // check the maximum product in // diagonal going through up - right if (($i - 3) >= 0 and ($j - 1) <= 0) { $result = $arr[$i][$j] * $arr[$i - 1][$j + 1] * $arr[$i - 2][$j + 2] * $arr[$i - 3][$j + 3]; if ($max < $result) $max = $result; } } } return $max; } // Driver Code $arr = array(array(1, 2, 3, 4, 5), array(6, 7, 8, 9, 1), array(2, 3, 4, 5, 6), array(7, 8, 9, 1, 0), array(9, 6, 4, 2, 3)); echo FindMaxProduct($arr, $n); // This code is contributed by anuj_67\. ?> ``` **输出**： ``` 3024 ``` * * * * * *