剑指offer十九之顺时针打印矩阵

public class PrintMatrix {
    public static void main(String[] args) {
        int[][] num = new int[][]{{1, 2, 3, 4, 5}, {6, 7, 8, 9, 10}, {11, 12, 13, 14, 15}, {16, 17, 18, 19, 20}, {21, 22, 23, 24, 25}};
        System.out.println("顺时针打印5x5的矩阵");
        printMatrixClockwisely(num, 5, 5);
        int[][] num2 = new int[][]{{1, 2, 3, 4, 5}};
        System.out.println("顺时针打印1x5的矩阵");
        printMatrixClockwisely(num2, 5, 1);
        int[][] num3 = new int[][]{{1}, {2}, {3}, {4}, {5}};
        System.out.println("顺时针打印5x1的矩阵");
        printMatrixClockwisely(num3, 1, 5);
    }

    //循环打印矩阵
    public static void printMatrixClockwisely(int[][] numbers, int clumns, int rows) {
        if (numbers == null || clumns <= 0 || rows <= 0)
            return;
        int start = 0;
        //对一个5*5的矩阵而言，最后一圈只有一个数字，对应的坐标为（2，2）。我们发现5>2*2.对于一个6*6的矩阵而言，最后一圈有4个数字，其左上角的坐标仍然为（2，2）。我们发现6>2*2依然成立。于是我们可以得出继续循环的条件是columns>2*start并且rows>2*start.
        while (clumns > 2 * start && rows > 2 * start) {
            printMatrixInCircle(numbers, clumns, rows, start);
            start++;
        }
    }

    //打印一圈
    private static void printMatrixInCircle(int[][] numbers, int clumns, int rows, int start) {
        int endX = clumns - 1 - start;
        int endY = rows - 1 - start;
        //从左到右打印一行
        for (int i = start; i <= endX; ++i) {
            int n = numbers[start][i];
            System.out.print(n + "   ");
        }
        System.out.println();
        //从上到下打印一列
        if (start < endY) {
            for (int j = start + 1; j <= endY; ++j) {
                int n = numbers[j][endX];
                System.out.print(n + "   ");
            }
        }
        System.out.println();
        //从右到左打印一行
        if (start < endY && start < endX) {
            for (int k = endX - 1; k >= start; k--) {
                int n = numbers[endY][k];
                System.out.print(n + "   ");
            }
        }
        System.out.println();
        //从下到上打印一列
        if (start < endX && start < endY - 1) {
            for (int l = endY - 1; l > start; l--) {
                int n = numbers[l][start];
                System.out.print(n + "   ");
            }
        }
        System.out.println();
    }
}