顺时针打印给定矩阵的边界元素
给定一个尺寸为 N * M 的矩阵 arr[][] ，任务是以顺时针形式打印给定矩阵的边界元素。
示例:
输入: arr[][] = {{1，2，3}，{4，5，6}，{7，8，9} } 输出: 1 2 3 6 9 8 7 4 说明: 矩阵的边界元素有: 1 2 3 T13】456 T18<强
输入: arr[][] = {{11，12，33}，{64，57，61}，{74，88，39}} 输出: 11 12 33 61 39 88 74 64
天真法:解决这个问题最简单的方法是遍历给定矩阵检查当前元素是否为边界元素。如果发现为真，则打印该元素。
时间复杂度:O(N2) 辅助空间: O(1)
高效途径:优化上述途径，思路是遍历只遍历矩阵的首末行和首末列。按照以下步骤解决问题:
- 打印矩阵的第一行。
- 打印矩阵除第一行以外的最后一列。
- 打印矩阵的最后一行，除了最后一列。
- 打印矩阵的第一列，第一行和最后一行除外。
下面是上述方法的实现:
C++
// C++ program of the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to print the boundary elements
// of the matrix in clockwise
void boundaryTraversal(vector<vector<int> > arr, int N,
int M)
{
// Print the first row
for (int i = 0; i < M; i++)
{
cout << arr[0][i] << " ";
}
// Print the last column
// except the first row
for (int i = 1; i < N; i++)
{
cout << arr[i][M - 1] << " ";
}
// Print the last row
// except the last column
if (N > 1)
{
// Print the last row
for (int i = M - 2; i >= 0; i--)
{
cout << arr[N - 1][i] << " ";
}
}
// Print the first column except
// the first and last row
if (M > 1) {
// Print the first column
for (int i = N - 2; i > 0; i--) {
cout << arr[i][0] << " ";
}
}
}
// Driver Code
int main()
{
vector<vector<int> > arr{ { 1, 2, 3 },
{ 4, 5, 6 },
{ 7, 8, 9 } };
int N = arr.size();
int M = arr[0].size();
// Function Call
boundaryTraversal(arr, N, M);
return 0;
}
// This code is contributed by Dharanendra L V
Java 语言(一种计算机语言，尤用于创建网站)
// Java program of the above approach
import java.util.*;
class GFG {
// Function to print the boundary elements
// of the matrix in clockwise
public static void boundaryTraversal(
int arr[][], int N, int M)
{
// Print the first row
for (int i = 0; i < M; i++) {
System.out.print(arr[0][i] + " ");
}
// Print the last column
// except the first row
for (int i = 1; i < N; i++) {
System.out.print(arr[i][M - 1] + " ");
}
// Print the last row
// except the last column
if (N > 1) {
// Print the last row
for (int i = M - 2; i >= 0; i--) {
System.out.print(arr[N - 1][i] + " ");
}
}
// Print the first column except
// the first and last row
if (M > 1) {
// Print the first column
for (int i = N - 2; i > 0; i--) {
System.out.print(arr[i][0] + " ");
}
}
}
// Driver Code
public static void main(String[] args)
{
int arr[][]
= { { 1, 2, 3 },
{ 4, 5, 6 },
{ 7, 8, 9 } };
int N = arr.length;
int M = arr[0].length;
// Function Call
boundaryTraversal(arr, N, M);
}
}
Python 3
# Python program of the above approach
# Function to print the boundary elements
# of the matrix in clockwise
def boundaryTraversal(arr, N, M):
# Print the first row
for i in range(M):
print(arr[0][i], end = " ");
# Print the last column
# except the first row
for i in range(1, N):
print(arr[i][M - 1], end = " ");
# Print the last row
# except the last column
if (N > 1):
# Print the last row
for i in range(M - 2, -1, -1):
print(arr[N - 1][i], end = " ");
# Print the first column except
# the first and last row
if (M > 1):
# Print the first column
for i in range(N - 2, 0, -1):
print(arr[i][0], end = " ");
# Driver Code
if __name__ == '__main__':
arr = [[1, 2, 3],
[4, 5, 6],
[7, 8, 9]];
N = len(arr);
M = len(arr[0]);
# Function Call
boundaryTraversal(arr, N, M);
# This code is contributed by 29AjayKumar
C
// C# program of the above approach
using System;
class GFG{
// Function to print the boundary elements
// of the matrix in clockwise
static void boundaryTraversal(int[,] arr,
int N, int M)
{
// Print the first row
for(int i = 0; i < M; i++)
{
Console.Write(arr[0, i] + " ");
}
// Print the last column
// except the first row
for(int i = 1; i < N; i++)
{
Console.Write(arr[i, M - 1] + " ");
}
// Print the last row
// except the last column
if (N > 1)
{
// Print the last row
for(int i = M - 2; i >= 0; i--)
{
Console.Write(arr[N - 1, i] + " ");
}
}
// Print the first column except
// the first and last row
if (M > 1)
{
// Print the first column
for(int i = N - 2; i > 0; i--)
{
Console.Write(arr[i, 0] + " ");
}
}
}
// Driver code
static void Main()
{
int[,] arr = { { 1, 2, 3 },
{ 4, 5, 6 },
{ 7, 8, 9 } };
int N = 3;
int M = 3;
// Function Call
boundaryTraversal(arr, N, M);
}
}
// This code is contributed by divyeshrabadiya07
java 描述语言
<script>
// Javascript program of the above approach
// Function to print the boundary elements
// of the matrix in clockwise
function boundaryTraversal(arr, N, M)
{
// Print the first row
for (let i = 0; i < M; i++)
{
document.write(arr[0][i] + " ");
}
// Print the last column
// except the first row
for (let i = 1; i < N; i++)
{
document.write(arr[i][M - 1] + " ");
}
// Print the last row
// except the last column
if (N > 1)
{
// Print the last row
for (let i = M - 2; i >= 0; i--)
{
document.write(arr[N - 1][i] + " ");
}
}
// Print the first column except
// the first and last row
if (M > 1) {
// Print the first column
for (let i = N - 2; i > 0; i--) {
document.write(arr[i][0] + " ");
}
}
}
// Driver Code
let arr = [ [ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ] ];
let N = arr.length;
let M = arr[0].length;
// Function Call
boundaryTraversal(arr, N, M);
</script>
Output:
1 2 3 6 9 8 7 4
时间复杂度: O(N + M) 辅助空间: O(1)
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