L形遍历矩阵

给定一个N * M矩阵。任务是以 L 形遍历给定矩阵，如下图所示。

例子：

Input : n = 3, m = 3
a[][] = { { 1, 2, 3 },
          { 4, 5, 6 },
          { 7, 8, 9 } } 
Output : 1 4 7 8 9 2 5 6 3

Input : n = 3, m = 4
a[][] = { { 1, 2, 3 },
          { 4, 5, 6 },
          { 7, 8, 9 },
          { 10, 11, 12} } 
Output : 1 4 7 10 11 12 2 5 8 9 3 6

观察将有 m（列数）个 L 形状需要遍历。因此，我们将分两部分遍历每个 L 形状，首先是垂直（从上到下），然后是水平（从左到右）。
要垂直遍历，观察每一列j ，0 <= j <= m – 1，我们需要垂直遍历 n – j 个元素。所以对于每一列 j，从 a[0][j] 遍历到 a[n-1-j][j]。
现在，要水平遍历每个 L 形状，观察每列 j 对应的行将是第 (n-1-j) 行，第一个元素将是从行开头的第 (j+1) 个元素。所以，对于每个 L 形状或每个列 j，要水平遍历，从 a[n-1-j][j+1] 遍历到 a[n-1-j][m-1]。
下面是这个方法的实现：

C++

// C++ program to traverse a m x n matrix in L shape.
#include 
using namespace std;
 
#define MAX 100
 
// Printing matrix in L shape
void traverseLshape(int a[][MAX], int n, int m)
{
    // for each column or each L shape
    for (int j = 0; j < m; j++) {
 
        // traversing vertically
        for (int i = 0; i <= n - j - 1; i++)
            cout << a[i][j] << " ";
 
        // traverse horizontally
        for (int k = j + 1; k < m; k++)
            cout << a[n - 1 - j][k] << " ";
    }
}
 
// Driven Program
int main()
{
    int n = 4;
    int m = 3;
    int a[][MAX] = { { 1, 2, 3 },
                     { 4, 5, 6 },
                     { 7, 8, 9 },
                     { 10, 11, 12 } };
    traverseLshape(a, n, m);
    return 0;
}

Java

// Java Program to traverse a m x n matrix in L shape.
public class GFG{
 
    static void traverseLshape(int a[][], int n, int m) {
        // for each column or each L shape
        for (int j = 0; j < m; j++) {
 
            // traversing vertically
            for (int i = 0; i <= n - j - 1; i++)
                System.out.print(a[i][j] + " ");
 
            // traverse horizontally
            for (int k = j + 1; k < m; k++)
                System.out.print(a[n - 1 - j][k] + " ");
        }
    }
 
    // Driver Code
    public static void main(String args[]) {
        int n = 4;
        int m = 3;
        int a[][] = { { 1, 2, 3 },
                        { 4, 5, 6 },
                        { 7, 8, 9 },
                        { 10, 11, 12 } };
        traverseLshape(a, n, m);
    }
}

Python3

# Python3 program to traverse a
# m x n matrix in L shape.
 
# Printing matrix in L shape
def traverseLshape(a, n, m):
     
    # for each column or each L shape
    for j in range(0, m):
 
        # traversing vertically
        for i in range(0, n - j):
            print(a[i][j], end = " ");
 
        # traverse horizontally
        for k in range(j + 1, m):
            print(a[n - 1 - j][k], end = " ");
 
# Driven Code
if __name__ == '__main__':
    n = 4;
    m = 3;
    a = [[1, 2, 3],
         [4, 5, 6],
         [7, 8, 9],
         [10, 11, 12]];
    traverseLshape(a, n, m);
 
# This code is contributed by PrinciRaj1992

C#

// C# Program to traverse a m x n matrix in L shape.
 
using System;
 
public class GFG{
   
    static void traverseLshape(int[,] a, int n, int m) {
        // for each column or each L shape
        for (int j = 0; j < m; j++) {
   
            // traversing vertically
            for (int i = 0; i <= n - j - 1; i++)
                Console.Write(a[i,j] + " ");
   
            // traverse horizontally
            for (int k = j + 1; k < m; k++)
                Console.Write(a[n - 1 - j,k] + " ");
        }
    }
   
    // Driver Code
    public static void Main() {
        int n = 4; 
        int m = 3; 
        int[,] a = { { 1, 2, 3 }, 
                        { 4, 5, 6 }, 
                        { 7, 8, 9 }, 
                        { 10, 11, 12 } }; 
        traverseLshape(a, n, m); 
    }
}

Javascript

输出：

1 4 7 10 11 12 2 5 8 9 3 6

时间复杂度： O(n * m)

辅助空间： O(1)