矩陣的邊界元素 · GeeksForGeeks 中文系列教程

# 矩陣的邊界元素 > 原文： [https://www.geeksforgeeks.org/boundary-elements-matrix/](https://www.geeksforgeeks.org/boundary-elements-matrix/) **打印矩陣的邊界元素。** 給定大小為 n x m 的矩陣。打印矩陣的邊界元素。邊界元素是在所有四個方向上都沒有被元素包圍的元素，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**：這個想法很簡單。遍歷矩陣并檢查每個元素是否在邊界上，如果是，則打印該元素，否則打印空格字符。 * **算法**： 1. 從頭到尾遍歷數組。 2. 分配外部循環以指向該行，分配內部行以遍歷該行的元素。 3. 如果元素位于矩陣的邊界中，則打印該元素，即元素是否位于第一行，第一列，最后一行，最后一列 4. 如果元素不是邊界元素，則打印空白。 * **實現**： ## 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 ?> ``` **輸出**： ``` 1 2 3 4 5 8 1 4 5 6 7 8 ``` * **復雜度分析**： * **時間復雜度**： O（n * n），其中 n 是數組的大小。這是通過矩陣的單次遍歷實現的。 * **空間復雜度**：`O(1)`。由于需要一個恒定的空間。 **查找邊界元素的總和** 給定大小為 n x m 的矩陣。找到矩陣邊界元素的總和。邊界元素是在所有四個方向上都未被元素包圍的元素，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**：這個想法很簡單。遍歷矩陣并檢查每個元素是否在邊界上，如果是，則將它們相加以獲得所有邊界元素的總和。 * **算法**： 1. 創建一個變量來存儲總和，并從頭到尾遍歷數組。 2. 分配外部循環以指向該行，分配內部行以遍歷該行的元素。 3. 如果元素位于矩陣的邊界中，則將第一個元素添加到總和中，即如果元素位于第一行，第一列，最后一行，最后一列 4. 打印總和。 * **實現**： ## 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 ?> ``` **輸出**： ``` Sum of boundary elements is 54 ``` * **復雜度分析**： * **時間復雜度**： O（n * n），其中 n 是數組的大小。這是通過矩陣的單次遍歷實現的。 * **空間復雜度**：`O(1)`。由于需要一個恒定的空間。