交换给定矩阵的上下三角一半

给定尺寸为N * N的正方形矩阵mat [] [] ，任务是打印在交换给定矩阵的上，下三角两半的横向倒置图像后可获得的矩阵。

Consider the matrix mat[][] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}
The lateral image of the lower triangular half of the matrix

The lateral image of the upper triangular half of the matrix

Therefore, following rearrangement of the matrix needs to be performed

例子：

Input: mat[][] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}
Output:
1  8  7
6  5  4
3  2  9
Explanation:

Input: mat[][] = {{1, 2}, {4, 5}}
Output:
1 4
2 5

方法：请按照以下步骤解决问题：

• 初始化向量数组upDiagonal和lowDiagonal ，分别存储来自下半三角形和上半三角形的矩阵元素的元素。
• 使用变量i和j分别遍历行和列遍历给定的矩阵，然后执行以下步骤：
  • 如果当前元素在主对角线上，则从此迭代继续。
  • 否则，如果当前元素存在于三角形的上半部分，则将该元素添加到索引为abs(i – j)的upDiagonal上。
  • 否则，将当前元素添加到索引为abs(i – j)的lowDiagonal上。
• 现在，再次穿过所述基质，并与来自下半和反之亦然的端部的元件替代任何元素存在于上半。
• 完成上述步骤后，打印所得矩阵。

下面是上述方法的实现：

C++

4
7 8

Java

6
 3 2

Python3

1  2  3    1  8  7
4  5  6 to 6  5  4
7  8  9    3  2  9

C#

1  2  3    6  5  4
1  8  7 to 7  8  9
4  5  6    3  2  9

输出：

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to swap laterally inverted
// images of upper and lower triangular
// halves of a given matrix
void ReverseSwap(vector >& mat, int n)
{
    // Store the matrix elements from
    // upper & lower triangular halves
    vector lowerEle[n];
    vector upperEle[n];
 
    int index;
 
    // Traverse the matrix mat[][]
    for (int i = 0; i < n; i++) {
 
        for (int j = 0; j < n; j++) {
 
            // Find the index
            index = abs(i - j);
 
            // If current element lies
            // on the principal diagonal
            if (i == j) {
                continue;
            }
 
            // If current element lies
            // below the principal diagonal
            else if (j < i) {
                lowerEle[index].push_back(
                    mat[i][j]);
            }
 
            // If current element lies
            // above the principal diagonal
            else {
                upperEle[index].push_back(
                    mat[i][j]);
            }
        }
    }
 
    // Traverse again to swap values
    for (int i = 0; i < n; i++) {
 
        for (int j = 0; j < n; j++) {
 
            // Find the index
            index = abs(i - j);
 
            // Principal diagonal
            if (i == j) {
                continue;
            }
 
            // Below main diagonal
            else if (j < i) {
 
                mat[i][j] = upperEle[index].back();
                upperEle[index].pop_back();
            }
 
            // Above main diagonal
            else {
                mat[i][j] = lowerEle[index].back();
                lowerEle[index].pop_back();
            }
        }
    }
 
    // Traverse the matrix and print
    for (int i = 0; i < n; i++) {
 
        for (int j = 0; j < n; j++) {
 
            cout << mat[i][j] << " ";
        }
        cout << endl;
    }
}
 
// Driver Code
int main()
{
    // Given Matrix mat[][]
    vector > mat = { { 1, 2 },
                                 { 4, 5 } };
 
    int N = mat.size();
 
    // Swap the upper and lower
    // triangular halves
    ReverseSwap(mat, N);
 
    return 0;
}

时间复杂度： O(N ² )
辅助空间： O(N ² )