leetcode-螺旋矩阵(指针）

给你一个 m 行 n 列的矩阵 matrix ，请按照 顺时针螺旋顺序 ，返回矩阵中的所有元素。

示例 1：

输入：matrix = [[1,2,3],[4,5,6],[7,8,9]]

输出：[1,2,3,6,9,8,7,4,5]

示例 2：

输入：matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]

输出：[1,2,3,4,8,12,11,10,9,5,6,7]

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int *spiralOrder(int **matrix, int matrixSize, int *matrixColSize, int *returnSize)
{

    int rows = matrixSize, columns = matrixColSize[0];
    int total = rows * columns;
    int *order = malloc(sizeof(int) * total);
    *returnSize = 0;
//  对变量进行抽象，可大大减小指针操作复杂度
    int left = 0, right = columns - 1, top = 0, bottom = rows - 1;
    while (left <= right && top <= bottom)
    {
        for (int column = left; column <= right; column++)
        {
            order[(*returnSize)++] = *(*(matrix + top) + column);
            printf("%d\n", *(*(matrix + top) + column));
        }

        for (int row = top + 1; row <= bottom; row++)
        {
            order[(*returnSize)++] = *(*(matrix + row) + right);
            printf("%d\n", *(*(matrix + row) + right));
        }
        if (left < right && top < bottom)
        {
            for (int column = right - 1; column > left; column--)
            {
                order[(*returnSize)++] = *(*(matrix + bottom) + column);

            }

            for (int row = bottom; row > top ; row--)
            {
                order[(*returnSize)++] = *(*(matrix + row) + left);

            }
        }
        left++;
        right--;
        top++;
        bottom--;
    }
    return order;
}