求矩阵所有边界和对角元素的和
给定顺序为 NxN 的 2D 数组 arr[][] ，任务是找到给定 arr[][] 的对角线和边界元素中存在的所有元素的总和。 示例:
输入: arr[][] = { {1，2，3，4}，{1，2，3，4}，{1，2，3，4}，{1，2，3，4} } 输出: 40 解释: 边界上的元素之和为 1+2+3+4+4+4+3+2+1+1 = 30。 对角线上不与边界元素相交的元素之和为 2 + 3 + 2 + 3 = 10。 因此所需的总和为 30 + 10 = 40。 输入: arr[][] = { {1，2，3}，{1，2，3}，{1，2，3}} 输出: 18 解释: 边界上的元素之和为 1+2+3+3+2+1+1 = 16。 对角线上不与边界元素相交的元素之和为 2。 因此所需的和为 16 + 2 = 18。
进场:
- 使用两个循环遍历给定的 2D 数组，一个循环用于行(比如说 i )，另一个循环用于列(比如说 j )。
- 如果 i 等于 j 或 (i + j) 等于(列-1 的大小)，则该元素对给定 2D 数组的对角线有贡献。
- 如果( i 或 j 等于 0 )或( i 或 j 等于第–1 列的 s 大小)，则该元素构成给定 2D 数组的边界元素。
- 满足上述两个条件的所有元素的和给出了所需的和。
以下是上述方法的实现:
C++
// C++ implementation of the above approach
#include "bits/stdc++.h"
using namespace std;
const int N = 4;
// Function to find the sum of all diagonal
// and Boundary elements
void diagonalBoundarySum(int arr[N][N])
{
int requiredSum = 0;
// Traverse arr[][]
// Loop from i to N-1 for rows
for (int i = 0; i < N; i++) {
// Loop from j = N-1 for columns
for (int j = 0; j < N; j++) {
// Condition for diagonal
// elements
if (i == j || (i + j) == N - 1) {
requiredSum += arr[i][j];
}
// Condition for Boundary
// elements
else if (i == 0 || j == 0
|| i == N - 1
|| j == N - 1) {
requiredSum += arr[i][j];
}
}
}
// Print the final Sum
cout << requiredSum << endl;
}
// Driver Code
int main()
{
int arr[][4] = { { 1, 2, 3, 4 },
{ 1, 2, 3, 4 },
{ 1, 2, 3, 4 },
{ 1, 2, 3, 4 } };
diagonalBoundarySum(arr);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the above approach
import java.util.*;
class GFG{
public static int N = 4;
// Function to find the sum of all diagonal
// and Boundary elements
static void diagonalBoundarySum(int arr[][]){
int requiredSum = 0;
// Traverse arr[][]
// Loop from i to N-1 for rows
for (int i = 0; i < N; i++) {
// Loop from j = N-1 for columns
for (int j = 0; j < N; j++) {
// Condition for diagonal
// elements
if (i == j || (i + j) == N - 1) {
requiredSum += arr[i][j];
}
// Condition for Boundary
// elements
else if (i == 0 || j == 0 || i == N - 1|| j == N - 1) {
requiredSum += arr[i][j];
}
}
}
// Print the final Sum
System.out.println(requiredSum);
}
// Driver Code
public static void main(String args[])
{
int arr[][] = { { 1, 2, 3, 4 },{ 1, 2, 3, 4 },
{ 1, 2, 3, 4 },{ 1, 2, 3, 4 } };
diagonalBoundarySum(arr);
}
}
// This code is contributed by AbhiThakur
Python 3
# Python implementation of the above approach
N = 4;
# Function to find the sum of all diagonal
# and Boundary elements
def diagonalBoundarySum(arr):
requiredSum = 0;
# Traverse arr
# Loop from i to N-1 for rows
for i in range(N):
# Loop from j = N-1 for columns
for j in range(N):
# Condition for diagonal
# elements
if (i == j or (i + j) == N - 1):
requiredSum += arr[i][j];
# Condition for Boundary
# elements
elif(i == 0 or j == 0 or i == N - 1 or j == N - 1):
requiredSum += arr[i][j];
# Print the final Sum
print(requiredSum);
# Driver Code
if __name__ == '__main__':
arr = [[ 1, 2, 3, 4 ],
[ 1, 2, 3, 4 ],
[ 1, 2, 3, 4 ],
[ 1, 2, 3, 4 ]];
diagonalBoundarySum(arr);
# This code is contributed by 29AjayKumar
C
// C# implementation of the above approach
using System;
class GFG
{
public static int N = 4;
// Function to find the sum of all diagonal
// and Boundary elements
static void diagonalBoundarySum(int[, ] arr){
int requiredSum = 0;
// Traverse arr[][]
// Loop from i to N-1 for rows
for (int i = 0; i < N; i++) {
// Loop from j = N-1 for columns
for (int j = 0; j < N; j++) {
// Condition for diagonal
// elements
if (i == j || (i + j) == N - 1) {
requiredSum += arr[i,j];
}
// Condition for Boundary
// elements
else if (i == 0 || j == 0 || i == N - 1|| j == N - 1) {
requiredSum += arr[i,j];
}
}
}
// Print the final Sum
Console.WriteLine(requiredSum);
}
// Driver Code
public static void Main()
{
int[, ] arr = { { 1, 2, 3, 4 },{ 1, 2, 3, 4 },{ 1, 2, 3, 4 },{ 1, 2, 3, 4 } };
diagonalBoundarySum(arr);
}
}
// This code is contributed by abhaysingh290895
java 描述语言
<script>
// Java script implementation of the above approach
let N = 4;
// Function to find the sum of all diagonal
// and Boundary elements
function diagonalBoundarySum(arr){
let requiredSum = 0;
// Traverse arr[][]
// Loop from i to N-1 for rows
for (let i = 0; i < N; i++) {
// Loop from j = N-1 for columns
for (let j = 0; j < N; j++) {
// Condition for diagonal
// elements
if (i == j || (i + j) == N - 1) {
requiredSum += arr[i][j];
}
// Condition for Boundary
// elements
else if (i == 0 || j == 0 || i == N - 1|| j == N - 1) {
requiredSum += arr[i][j];
}
}
}
// Print the final Sum
document.write(requiredSum);
}
// Driver Code
let arr = [[ 1, 2, 3, 4 ],[ 1, 2, 3, 4 ],
[1, 2, 3, 4 ],[ 1, 2, 3, 4 ]];
diagonalBoundarySum(arr);
// contributed by sravan kumar
</script>
Output:
40
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