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/
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。
-
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。 -
次对角线的条件:行列条件为
row = numberOfRows - column - 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++
// 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
// 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.
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;
}
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.
PHP
<?php
// An efficient 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++)
{
$principal += $mat[$i][$i];
$secondary += $mat[$i][$n - $i - 1];
}
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 contributed by aj_36
?>
输出:
Principal Diagonal:18
Secondary Diagonal:18
此代码需要O(n)
时间和O(1)
辅助空间。
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