矩阵的边界元素 · GeeksForGeeks 中文系列教程

# 矩阵的边界元素 > 原文： [https://www.geeksforgeeks.org/boundary-elements-matrix/](https://www.geeksforgeeks.org/boundary-elements-matrix/) **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。 * **算法**： 1. 从头到尾遍历数组。 2. 分配外部循环以指向该行，分配内部行以遍历该行的元素。 3. 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 4. 如果元素不是边界元素，则打印空白。 * **实现**： ## C++ ``` // C++ program to print boundary element of // matrix. #include <bits/stdc++.h> 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. ``` ## 蛇蟒 ``` # 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**： 1. 创建一个变量来存储总和，并从头到尾遍历数组。 2. 分配外部循环以指向该行，分配内部行以遍历该行的元素。 3. 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 4. 打印总和。 * **实现**： ## C++ ``` // C++ program to find sum of boundary elements // of matrix. #include <bits/stdc++.h> 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。