C ++程序来查找两条对角线之和之间的差异

给定一个n X n的矩阵。任务是计算其对角线之和之间的绝对差。
例子：

Input : mat[][] = 11 2 4
                   4 5 6
                  10 8 -12 
Output : 15
Sum of primary diagonal = 11 + 5 + (-12) = 4.
Sum of primary diagonal = 4 + 5 + 10 = 19.
Difference = |19 - 4| = 15.


Input : mat[][] = 10 2
                   4 5
Output : 7

计算一个方阵的两条对角线的和。沿着矩阵的第一条对角线，行索引 = 列索引，即如果 i = j，则 mat[i][j] 位于第一条对角线上。沿着另一条对角线，如果 i = n-1-j，行索引 = n – 1 – 列索引，即 mat[i][j] 位于第二条对角线上。通过使用两个循环，我们遍历整个矩阵并计算矩阵对角线的总和。
下面是这种方法的实现：

C++

// C++ program to find the difference
// between the sum of diagonal.
#include 
#define MAX 100
using namespace std;
  
int difference(int arr[][MAX], int n)
{
    // Initialize sums of diagonals
    int d1 = 0, d2 = 0;
  
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < n; j++)
        {
            // finding sum of primary diagonal
            if (i == j)
                d1 += arr[i][j];
  
            // finding sum of secondary diagonal
            if (i == n - j - 1)
                d2 += arr[i][j];
        }
    }
  
    // Absolute difference of the sums
    // across the diagonals
    return abs(d1 - d2);
}
  
// Driven Program
int main()
{
    int n = 3;
  
    int arr[][MAX] =
    {
        {11, 2, 4},
        {4 , 5, 6},
        {10, 8, -12}
    };
  
    cout << difference(arr, n);
    return 0;
}

C++

// C++ program to find the difference
// between the sum of diagonal.
#include 
#define MAX 100
using namespace std;
  
int difference(int arr[][MAX], int n)
{
    // Initialize sums of diagonals
    int d1 = 0, d2 = 0;
  
    for (int i = 0; i < n; i++)
    {
        d1 += arr[i][i];
        d2 += arr[i][n-i-1];
    }
  
    // Absolute difference of the sums
    // across the diagonals
    return abs(d1 - d2);
}
  
// Driven Program
int main()
{
    int n = 3;
  
    int arr[][MAX] =
    {
        {11, 2, 4},
        {4 , 5, 6},
        {10, 8, -12}
    };
  
    cout << difference(arr, n);
    return 0;
}

输出：

时间复杂度： O(n*n)
我们可以使用单元格索引中存在的模式优化上述解决方案以在 O(n) 中工作。

C++

// C++ program to find the difference
// between the sum of diagonal.
#include 
#define MAX 100
using namespace std;
  
int difference(int arr[][MAX], int n)
{
    // Initialize sums of diagonals
    int d1 = 0, d2 = 0;
  
    for (int i = 0; i < n; i++)
    {
        d1 += arr[i][i];
        d2 += arr[i][n-i-1];
    }
  
    // Absolute difference of the sums
    // across the diagonals
    return abs(d1 - d2);
}
  
// Driven Program
int main()
{
    int n = 3;
  
    int arr[][MAX] =
    {
        {11, 2, 4},
        {4 , 5, 6},
        {10, 8, -12}
    };
  
    cout << difference(arr, n);
    return 0;
}

输出：

时间复杂度： O(n)
有关详细信息，请参阅有关查找两条对角线之和之间的差异的完整文章！