有效地计算矩阵的对角线之和

给定一个二维方阵，求主对角线和次对角线中元素的总和。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. 主对角线条件：行列条件是行 = 列。
   次对角线由元素 A03、A12、A21、A30 形成。
2. 次对角线的条件：行列条件是 row = numberOfRows – column -1。

例子 ：

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) ：

在这种方法中，我们使用两个循环，即一个循环用于列和一个循环用于行，并且在内部循环中，我们检查上述条件：

C++

// A simple C++ program to find sum of diagonals
#include 
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

// 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

# 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#

// 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.

Javascript

// A simple Javascript program to find sum of diagonals
 
const MAX = 100;
 
void printDiagonalSums(mat, n)
{
    let principal = 0, secondary = 0;
    for (let i = 0; i < n; i++) {
        for (let 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];
        }
    }
 
    document.write("Principal Diagonal:" + principal + "<br>");
    document.write("Secondary Diagonal:" + secondary + "<br>");
}
 
// Driver code
    let 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 subhammahato348.

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）：

在这种方法中，我们使用一个循环，即一个用于计算主对角线和次对角线之和的循环：

C++

// An efficient C++ program to find sum of diagonals
#include 
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;
}

Java

// An efficient 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++) {
            principal += mat[i][i];
            secondary += mat[i][n - i - 1];
        }
     
        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

# A simple Python3 program to find
# sum of diagonals
MAX = 100
 
def printDiagonalSums(mat, n):
 
    principal = 0
    secondary = 0
    for i in range(0, n):
        principal += mat[i][i]
        secondary += mat[i][n - i - 1]
         
    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＃

// An efficient 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++) {
            principal += mat[i,i];
            secondary += mat[i,n - i - 1];
        }
     
        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.

Javascript

 
// An efficient Javascript  program to find
// sum of diagonals
 
function  printDiagonalSums(mat,n)
    {
        let principal = 0, secondary = 0;
        for (let i = 0; i < n; i++) {
            principal += mat[i][i];
            secondary += mat[i][n - i - 1];
        }
     
        document.write("Principal Diagonal:"
                                + principal+"<br>");
                                     
        document.write("Secondary Diagonal:"
                                + secondary);
    }
     
    // Driver code
     
        let a = [[ 1, 2, 3, 4 ],
                    [5, 6, 7, 8 ],
                    [ 1, 2, 3, 4 ],
                    [ 5, 6, 7, 8 ]];
     
        printDiagonalSums(a, 4);
         
// This code is contributed Bobby
 

输出 ：

Principal Diagonal:18
Secondary Diagonal:18