矩阵的边界元素
打印矩阵的边界元素。
给定大小为n x m的矩阵。 打印矩阵的边界元素。 边界元素是在所有四个方向上都没有被元素包围的元素，即第一行，第一列，最后一行和最后一列中的元素。
示例：
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在边界上，如果是，则打印该元素，否则打印空格字符。
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算法：
-
从头到尾遍历数组。
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分配外部循环以指向该行，分配内部行以遍历该行的元素。
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如果元素位于矩阵的边界中，则打印该元素，即元素是否位于第一行，第一列，最后一行，最后一列。
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如果元素不是边界元素，则打印空白。
-
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实现：
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
```
PHP
```
<?php // PHP program to print // boundary element of // matrix. $MAX = 100;
function printBoundary($a, $m, $n) { global $MAX; for ($i = 0; $i < $m; $i++) { for ($j = 0; $j < $n; $j++) { if ($i == 0) echo $a[$i][$j], " "; else if ($i == $m - 1) echo $a[$i][$j], " "; else if ($j == 0) echo $a[$i][$j], " "; else if ($j == $n - 1) echo $a[$i][$j], " "; else echo " ", " "; } echo "\n"; } }
// Driver code $a = array(array( 1, 2, 3, 4 ), array( 5, 6, 7, 8 ), array( 1, 2, 3, 4 ), array( 5, 6, 7, 8 )); printBoundary($a, 4, 4);
// This code is contributed // by akt_mit ?>
```
输出：
``` 1 2 3 4 5 8 1 4 5 6 7 8
```
-
复杂度分析：
-
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实现的。
-
空间复杂度：
O(1)。由于需要一个恒定的空间。
-
查找边界元素的总和：
给定大小为n x m的矩阵。 找到矩阵边界元素的总和。 边界元素是在所有四个方向上都未被元素包围的元素，即第一行，第一列，最后一行和最后一列中的元素。
示例：
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在边界上，如果是，则将它们相加以获得所有边界元素的总和。
-
算法：
-
创建一个变量来存储总和，并从头到尾遍历数组。
-
分配外部循环以指向该行，分配内部行以遍历该行的元素。
-
如果元素位于矩阵的边界中，则将第一个元素添加到总和中，即如果元素位于第一行，第一列，最后一行，最后一列。
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打印总和。
-
-
实现：
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
```
PHP
```
<?php // PHP program to find // sum of boundary // elements of matrix.
function getBoundarySum($a, $m, $n) { $sum = 0; for ($i = 0; $i < $m; $i++) { for ( $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 $a = array(array(1, 2, 3, 4), array(5, 6, 7, 8), array(1, 2, 3, 4), array(5, 6, 7, 8));
$sum = getBoundarySum($a, 4, 4); echo "Sum of boundary elements is ", $sum;
// This code is contributed by ajit ?>
```
输出：
``` Sum of boundary elements is 54
```
-
复杂度分析：
-
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实现的。
-
空间复杂度：
O(1)。由于需要一个恒定的空间。
-
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