矩阵的边界元素

打印矩阵的边界元素。

给定一个大小为 nx m 的矩阵。打印矩阵的边界元素。边界元素是那些没有被所有四个方向上的元素包围的元素，即第一行、第一列、最后一行和最后一列中的元素。

例子：

Input :
        1 2 3 4  
        5 6 7 8
        1 2 3 4
        5 6 7 8
Output : 
         1 2 3 4 
         5     8 
         1     4 
         5 6 7 8
Explanation:The boundary elements of the
matrix is printed.

Input:
        1 2 3   
        5 6 7 
        1 2 3 
Output: 
        1 2 3   
        5   7 
        1 2 3 
Explanation:The boundary elements of the 
matrix is printed.

方法：这个想法很简单。遍历矩阵并检查每个元素是否位于边界上，如果是，则打印元素，否则打印空格字符。

• 算法 ：
  1. 从头到尾遍历数组。
  2. 分配外循环指向行，内行遍历行的元素。
  3. 如果元素位于矩阵的边界，则打印该元素，即如果元素位于第一行，第一列，最后一行，最后一列
  4. 如果元素不是边界元素，则打印一个空格。
• 执行：

C++

// C++ program to print boundary element of
// matrix.
#include 
using namespace std;
 
const int MAX = 100;
 
void printBoundary(int a[][MAX], int m, int n)
{
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            if (i == 0 || j == 0 || i == n - 1 || j == n - 1)
                cout << a[i][j] << " ";
            else
                cout << " "
                     << " ";
        }
        cout << "\n";
    }
}
 
// Driver code
int main()
{
    int a[4][MAX] = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 1, 2, 3, 4 }, { 5, 6, 7, 8 } };
    printBoundary(a, 4, 4);
    return 0;
}

Java

// JAVA Code for Boundary elements of a Matrix
class GFG {
 
    public static void printBoundary(int a[][], int m,
                                     int n)
    {
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (i == 0)
                    System.out.print(a[i][j] + " ");
                else if (i == m - 1)
                    System.out.print(a[i][j] + " ");
                else if (j == 0)
                    System.out.print(a[i][j] + " ");
                else if (j == n - 1)
                    System.out.print(a[i][j] + " ");
                else
                    System.out.print("  ");
            }
            System.out.println("");
        }
    }
 
    /* Driver program to test above function */
    public static void main(String[] args)
    {
        int a[][] = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 1, 2, 3, 4 }, { 5, 6, 7, 8 } };
 
        printBoundary(a, 4, 4);
    }
}
// This code is contributed by Arnav Kr. Mandal.

Python

# Python program to print boundary element
# of the matrix.
 
MAX = 100
  
def printBoundary(a, m, n):
    for i in range(m):
        for j in range(n):
            if (i == 0):
                print a[i][j],
            elif (i == m-1):
                print a[i][j],
            elif (j == 0):
                print a[i][j],
            elif (j == n-1):
                print a[i][j],
            else:
                print " ",
        print
         
# Driver code
a = [ [ 1, 2, 3, 4 ], [ 5, 6, 7, 8 ],
    [ 1, 2, 3, 4 ], [ 5, 6, 7, 8 ] ]
printBoundary(a, 4, 4)
 
# This code is contributed by Sachin Bisht

C#

// C# Code for Boundary
// elements of a Matrix
using System;
 
class GFG {
 
    public static void printBoundary(int[, ] a,
                                     int m,
                                     int n)
    {
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (i == 0)
                    Console.Write(a[i, j] + " ");
                else if (i == m - 1)
                    Console.Write(a[i, j] + " ");
                else if (j == 0)
                    Console.Write(a[i, j] + " ");
                else if (j == n - 1)
                    Console.Write(a[i, j] + " ");
                else
                    Console.Write("  ");
            }
            Console.WriteLine(" ");
        }
    }
 
    // Driver Code
    static public void Main()
    {
        int[, ] a = { { 1, 2, 3, 4 },
                      { 5, 6, 7, 8 },
                      { 1, 2, 3, 4 },
                      { 5, 6, 7, 8 } };
 
        printBoundary(a, 4, 4);
    }
}
 
// This code is contributed by ajit

Javascript

   
// JavaScript Code for Boundary
// elements of a Matrix
function printBoundary(a, m, n)
{
    for (var i = 0; i < m; i++) {
        for (var j = 0; j < n; j++) {
            if (i == 0)
                document.write(a[i][j] + '\xa0');
            else if (i == m - 1)
                document.write(a[i][j] + '\xa0');
            else if (j == 0)
                document.write(a[i][j] + '\xa0');
            else if (j == n - 1)
                document.write(a[i][j] + '\xa0');
            else
                document.write("\xa0\xa0\xa0");
        }
        document.write("\xa0<br>");
    }
}
// Driver Code
var a = [ [ 1, 2, 3, 4 ],
              [ 5, 6, 7, 8 ],
              [ 1, 2, 3, 4 ],
              [ 5, 6, 7, 8 ]];
printBoundary(a, 4, 4);
 
 

输出：

1 2 3 4 
5     8 
1     4 
5 6 7 8

• 复杂性分析：
  • 时间复杂度： O(n*n)，其中 n 是数组的大小。
    这是通过矩阵的单次遍历来实现的。
  • 空间复杂度： O(1)。
    因为需要一个恒定的空间。

找到边界元素的总和

给定一个大小为 nx m 的矩阵。求矩阵的边界元素之和。边界元素是那些没有被所有四个方向上的元素包围的元素，即第一行、第一列、最后一行和最后一列中的元素。

例子：

Input :
        1 2 3 4  
        5 6 7 8
        1 2 3 4
        5 6 7 8
Output : 54
Explanation:The boundary elements of the matrix 
         1 2 3 4 
         5     8 
         1     4 
         5 6 7 8
Sum = 1+2+3+4+5+8+1+4+5+6+7+8 =54

Input :
        1 2 3   
        5 6 7 
        1 2 3 
Output : 24
Explanation:The boundary elements of the matrix
        1 2 3   
        5   7 
        1 2 3  
Sum = 1+2+3+5+7+1+2+3 = 24

方法：这个想法很简单。遍历矩阵并检查每个元素是否位于边界上，如果是，则将它们相加以获得所有边界元素的总和。

• 算法 ：
  1. 创建一个变量来存储总和并从头到尾遍历数组。
  2. 分配外循环指向行，内行遍历行的元素。
  3. 如果元素位于矩阵的边界，则将该元素添加到总和中，即如果元素位于第一行，第一列，最后一行，最后一列
  4. 打印总和。
• 执行：

C++

// C++ program to find sum of boundary elements
// of matrix.
#include 
using namespace std;
 
const int MAX = 100;
 
int getBoundarySum(int a[][MAX], int m, int n)
{
    long long int sum = 0;
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            if (i == 0)
                sum += a[i][j];
            else if (i == m - 1)
                sum += a[i][j];
            else if (j == 0)
                sum += a[i][j];
            else if (j == n - 1)
                sum += a[i][j];
        }
    }
    return sum;
}
 
// Driver code
int main()
{
    int a[][MAX] = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 1, 2, 3, 4 }, { 5, 6, 7, 8 } };
    long long int sum = getBoundarySum(a, 4, 4);
    cout << "Sum of boundary elements is " << sum;
    return 0;
}

Java

// JAVA Code for Finding sum of boundary elements
class GFG {
 
    public static long getBoundarySum(int a[][], int m,
                                      int n)
    {
        long sum = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (i == 0)
                    sum += a[i][j];
                else if (i == m - 1)
                    sum += a[i][j];
                else if (j == 0)
                    sum += a[i][j];
                else if (j == n - 1)
                    sum += a[i][j];
            }
        }
        return sum;
    }
 
    /* Driver program to test above function */
    public static void main(String[] args)
    {
        int a[][] = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 1, 2, 3, 4 }, { 5, 6, 7, 8 } };
        long sum = getBoundarySum(a, 4, 4);
        System.out.println("Sum of boundary elements"
                           + " is " + sum);
    }
}
// This code is contributed by Arnav Kr. Mandal.

Python

# Python program to print boundary element
# of the matrix.
 
MAX = 100
  
def printBoundary(a, m, n):
    sum = 0
    for i in range(m):
        for j in range(n):
            if (i == 0):
                sum += a[i][j]
            elif (i == m-1):
                sum += a[i][j]
            elif (j == 0):
                sum += a[i][j]
            elif (j == n-1):
                sum += a[i][j]
    return sum
     
# Driver code
a = [ [ 1, 2, 3, 4 ], [ 5, 6, 7, 8 ],
    [ 1, 2, 3, 4 ], [ 5, 6, 7, 8 ] ]
sum = printBoundary(a, 4, 4)
print "Sum of boundary elements is", sum
 
# This code is contributed by Sachin Bisht

C＃

// C# Code for Finding sum
// of boundary elements
using System;
 
class GFG {
    public static long getBoundarySum(int[, ] a,
                                      int m, int n)
    {
        long sum = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (i == 0)
                    sum += a[i, j];
                else if (i == m - 1)
                    sum += a[i, j];
                else if (j == 0)
                    sum += a[i, j];
                else if (j == n - 1)
                    sum += a[i, j];
            }
        }
        return sum;
    }
 
    // Driver Code
    static public void Main()
    {
        int[, ] a = { { 1, 2, 3, 4 },
                      { 5, 6, 7, 8 },
                      { 1, 2, 3, 4 },
                      { 5, 6, 7, 8 } };
        long sum = getBoundarySum(a, 4, 4);
        Console.WriteLine("Sum of boundary"
                          + " elements is " + sum);
    }
}
 
// This code is contributed by ajit

Javascript

 
// Javascript code for finding sum
// of boundary elements
function getBoundarySum(a, m, n)
{
    let sum = 0;
    for(let i = 0; i < m; i++)
    {
        for(let j = 0; j < n; j++)
        {
            if (i == 0)
                sum += a[i][j];
            else if (i == m - 1)
                sum += a[i][j];
            else if (j == 0)
                sum += a[i][j];
            else if (j == n - 1)
                sum += a[i][j];
        }
    }
    return sum;
}
 
// Driver code
let a = [ [ 1, 2, 3, 4 ],
          [ 5, 6, 7, 8 ],
          [ 1, 2, 3, 4 ],
          [ 5, 6, 7, 8 ] ];
let sum = getBoundarySum(a, 4, 4);
 
document.write("Sum of boundary elements" +
               " is " + sum);
                
// This code is contributed by rameshtravel07 
 

输出：

Sum of boundary elements is 54

• 复杂性分析：
  • 时间复杂度： O(n*n)，其中 n 是数组的大小。
    这是通过矩阵的单次遍历来实现的。
  • 空间复杂度： O(1)。
    因为需要一个恒定的空间。