找出两个对角线之和之间的差
原文： https://www.geeksforgeeks.org/find-difference-between-sums-of-two-diagonals/
给定n x n的矩阵。 任务是计算对角线总和之间的绝对差。
示例：
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
计算方阵的两个对角线的和。 沿着矩阵的第一个对角线，行索引等于列索引，即如果i = j，则mat[i][j]位于第一个对角线上。 沿着另一个对角线，i = n – 1 – j，即如果i = n - 1 - j，则mat[i][j]位于第二对角线上。 通过使用两个循环，我们遍历整个矩阵并计算矩阵对角线上的总和。
以下是此方法的实现：
C++
// C++ program to find the difference
// between the sum of diagonal.
#include <bits/stdc++.h>
#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;
}
Java
// JAVA Code for Find difference between sums
// of two diagonals
class GFG {
public static int difference(int arr[][], 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 Math.abs(d1 - d2);
}
/* Driver program to test above function */
public static void main(String[] args)
{
int n = 3;
int arr[][] =
{
{11, 2, 4},
{4 , 5, 6},
{10, 8, -12}
};
System.out.print(difference(arr, n));
}
}
// This code is contributed by Arnav Kr. Mandal.
Python3
# Python3 program to find the difference
# between the sum of diagonal.
def difference(arr, n):
# Initialize sums of diagonals
d1 = 0
d2 = 0
for i in range(0, n):
for j in range(0, n):
# 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);
# Driver Code
n = 3
arr = [[11, 2, 4],
[4 , 5, 6],
[10, 8, -12]]
print(difference(arr, n))
# This code is contributed
# by ihritik
C
// C# Code for find difference between
// sums of two diagonals
using System;
public class GFG
{
// Function to calculate difference
public static int difference(int[,] arr,
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 Math.Abs(d1 - d2);
}
// Driver Code
public static void Main()
{
int n = 3;
int[,] arr ={{11, 2, 4},
{4 , 5, 6},
{10, 8, -12}};
Console.Write(difference(arr, n));
}
}
// This code is contributed by shiv_bhakt.
PHP
<?php
// PHP program to find the difference
// between the sum of diagonal.
function difference($arr, $n)
{
// Initialize sums of diagonals
$d1 = 0; $d2 = 0;
for ($i = 0; $i < $n; $i++)
{
for ($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);
}
// Driver Code
{
$n = 3;
$arr = array(array(11, 2, 4),
array(4 , 5, 6),
array(10, 8, -12));
echo difference($arr, $n);
return 0;
}
// This code is contributed by nitin mittal.
?>
输出：
15
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来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 <bits/stdc++.h>
#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;
}
Java
// JAVA Code for Find difference between sums
// of two diagonals
class GFG {
public static int difference(int arr[][], 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 Math.abs(d1 - d2);
}
/* Driver program to test above function */
public static void main(String[] args)
{
int n = 3;
int arr[][] =
{
{11, 2, 4},
{4 , 5, 6},
{10, 8, -12}
};
System.out.print(difference(arr, n));
}
}
// This code is contributed by Arnav Kr. Mandal.
Python3
# Python3 program to find the difference
# between the sum of diagonal.
def difference(arr, n):
# Initialize sums of diagonals
d1 = 0
d2 = 0
for i in range(0, n):
d1 = d1 + arr[i][i]
d2 = d2 + arr[i][n - i - 1]
# Absolute difference of the sums
# across the diagonals
return abs(d1 - d2)
# Driver Code
n = 3
arr = [[11, 2, 4],
[4 , 5, 6],
[10, 8, -12]]
print(difference(arr, n))
# This code is contributed
# by ihritik
C
// C# Code for find difference between
// sums of two diagonals
using System;
public class GFG
{
//Function to find difference
public static int difference(int[,] arr,
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 Math.Abs(d1 - d2);
}
// Driver Code
public static void Main()
{
int n = 3;
int[,] arr ={{11, 2, 4},
{4 , 5, 6},
{10, 8, -12}};
Console.Write(difference(arr, n));
}
}
// This code is contributed by shiv_bhakt.
PHP
<?php
// PHP program to find the difference
// between the sum of diagonal.
function difference($arr, $n)
{
// Initialize sums of diagonals
$d1 = 0; $d2 = 0;
for ($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);
}
// Driver Code
{
$n = 3;
$arr =array(array(11, 2, 4),
array(4, 5, 6),
array(10, 8, -12));
echo difference($arr, $n);
return 0;
}
// This code is contributed by nitin mittal.
?>
输出：
15
时间复杂度：O(n)。
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