有效地计算矩阵的对角线总和 · GeeksForGeeks 中文系列教程

# 有效地计算矩阵的对角线总和 > 原文： [https://www.geeksforgeeks.org/efficiently-compute-sums-of-diagonals-of-a-matrix/](https://www.geeksforgeeks.org/efficiently-compute-sums-of-diagonals-of-a-matrix/) 给定一个 2D 正方形矩阵，请找到主对角线和辅助对角线中的元素之和。例如，考虑以下 4 X 4 输入矩阵。 ``` A00 A01 A02 A03 A10 A11 A12 A13 A20 A21 A22 A23 A30 A31 A32 A33 ``` 主对角线由元素 A00，A11，A22，A33 形成。 1. 主对角线的条件：行-列条件为 row = column。次对角线由元素 A03，A12，A21，A30 形成。 2. 次对角线的条件：行-列条件为 row = numberOfRows –列-1。 **示例**： ``` Input : 4 1 2 3 4 4 3 2 1 7 8 9 6 6 5 4 3 Output : Principal Diagonal: 16 Secondary Diagonal: 20 Input : 3 1 1 1 1 1 1 1 1 1 Output : Principal Diagonal: 3 Secondary Diagonal: 3 ``` **方法 1（O（n ^ 2）**：在此方法中，我们使用两个循环，即列循环和行循环，在内部循环中我们检查上述条件： ## C++ ```cpp // A simple C++ program to find sum of diagonals #include <bits/stdc++.h> using namespace std; const int MAX = 100; void printDiagonalSums(int mat[][MAX], int n) { int principal = 0, secondary = 0; for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { // Condition for principal diagonal if (i == j) principal += mat[i][j]; // Condition for secondary diagonal if ((i + j) == (n - 1)) secondary += mat[i][j]; } } cout << "Principal Diagonal:" << principal << endl; cout << "Secondary Diagonal:" << secondary << endl; } // Driver code int main() { int a[][MAX] = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 1, 2, 3, 4 }, { 5, 6, 7, 8 } }; printDiagonalSums(a, 4); return 0; } ``` ## Java ```java // A simple java program to find // sum of diagonals import java.io.*; public class GFG { static void printDiagonalSums(int [][]mat, int n) { int principal = 0, secondary = 0; for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { // Condition for principal // diagonal if (i == j) principal += mat[i][j]; // Condition for secondary // diagonal if ((i + j) == (n - 1)) secondary += mat[i][j]; } } System.out.println("Principal Diagonal:" + principal); System.out.println("Secondary Diagonal:" + secondary); } // Driver code static public void main (String[] args) { int [][]a = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 1, 2, 3, 4 }, { 5, 6, 7, 8 } }; printDiagonalSums(a, 4); } } // This code is contributed by vt_m. ``` ## Python3 ```py # A simple Python program to # find sum of diagonals MAX = 100 def printDiagonalSums(mat, n): principal = 0 secondary = 0; for i in range(0, n): for j in range(0, n): # Condition for principal diagonal if (i == j): principal += mat[i][j] # Condition for secondary diagonal if ((i + j) == (n - 1)): secondary += mat[i][j] print("Principal Diagonal:", principal) print("Secondary Diagonal:", secondary) # Driver code a = [[ 1, 2, 3, 4 ], [ 5, 6, 7, 8 ], [ 1, 2, 3, 4 ], [ 5, 6, 7, 8 ]] printDiagonalSums(a, 4) # This code is contributed # by ihritik ``` ## C# ```cs // A simple C# program to find sum // of diagonals using System; public class GFG { static void printDiagonalSums(int [,]mat, int n) { int principal = 0, secondary = 0; for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { // Condition for principal // diagonal if (i == j) principal += mat[i,j]; // Condition for secondary // diagonal if ((i + j) == (n - 1)) secondary += mat[i,j]; } } Console.WriteLine("Principal Diagonal:" + principal); Console.WriteLine("Secondary Diagonal:" + secondary); } // Driver code static public void Main () { int [,]a = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 1, 2, 3, 4 }, { 5, 6, 7, 8 } }; printDiagonalSums(a, 4); } } // This code is contributed by vt_m. ``` ## PHP ```php <?php // A simple PHP program to // find sum of diagonals $MAX = 100; function printDiagonalSums($mat, $n) { global $MAX; $principal = 0; $secondary = 0; for ($i = 0; $i < $n; $i++) { for ($j = 0; $j < $n; $j++) { // Condition for // principal diagonal if ($i == $j) $principal += $mat[$i][$j]; // Condition for // secondary diagonal if (($i + $j) == ($n - 1)) $secondary += $mat[$i][$j]; } } echo "Principal Diagonal:" , $principal ,"\n"; echo "Secondary Diagonal:", $secondary ,"\n"; } // Driver code $a = array (array ( 1, 2, 3, 4 ), array ( 5, 6, 7, 8 ), array ( 1, 2, 3, 4 ), array ( 5, 6, 7, 8 )); printDiagonalSums($a, 4); // This code is contrbuted by ajit ?> ``` **输出**： ``` Principal Diagonal:18 Secondary Diagonal:18 ``` 此代码需要 O（n ^ 2）时间和`O(1)`辅助空间方法 2（`O(n)`：在此方法中，我们使用一个循环，即用于计算主对角线和辅助对角线之和的循环： ## C++ ``` // An efficient C++ program to find sum of diagonals #include <bits/stdc++.h> using namespace std; const int MAX = 100; void printDiagonalSums(int mat[][MAX], int n) { int principal = 0, secondary = 0; for (int i = 0; i < n; i++) { principal += mat[i][i]; secondary += mat[i][n - i - 1]; } cout << "Principal Diagonal:" << principal << endl; cout << "Secondary Diagonal:" << secondary << endl; } // Driver code int main() { int a[][MAX] = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 1, 2, 3, 4 }, { 5, 6, 7, 8 } }; printDiagonalSums(a, 4); return 0; } ```