转圈打印矩阵

题目：

思路：

本题难点：由于是整型矩阵，所以存在5*5,4*5,4*6，这种，可能出现还剩下一列或者一行或者单个的情况。

代码示例：

import java.util.Arrays;

public class Solution2 {

    public static void main(String[] args) {

        int len = 7, higt = 10;

        int[][] A = new int[higt][len];

        for (int i = 0; i < higt; i++) {

            for (int j = 0; j < len; j++) {

                A[i][j] = j + 1 + i * len;

                System.out.print(A[i][j] + "\t");

            }

            System.out.println();

        }

        System.out.println(Arrays.toString(printMatrix(A)));

    }

    /**

     * @param matrix int整型二维数组 the matrix

     * @return int整型一维数组

     */

    public static int[] printMatrix(int[][] matrix) {

        int length = matrix[0].length, high = matrix.length;

        int num = length * high;

        int[] result = new int[num];

        int upper = 0, lower = high - 1, left = 0, right = length - 1;

        int index = 0;

        while (upper < lower && left < right) {

            for (int i = left; i <= right; i++) {

                result[index++] = matrix[upper][i];

            }

            upper++;

            for (int i = upper; i <= lower; i++) {

                result[index++] = matrix[i][right];

            }

            right--;

            for (int i = right; i >= left; i--) {

                result[index++] = matrix[lower][i];

            }

            lower--;

            for (int i = lower; i >= upper; i--) {

                result[index++] = matrix[i][left];

            }

            left++;

        }

        //由于是整型矩阵，所以存在5*5,4*5,4*6，这种，可能出现还剩下一列或者一行或者单个的情况

        if (left == right && upper < lower) {

            for (int i = upper; i <= lower; i++) result[index++] = matrix[i][left];

        } else if (upper == lower && left < right) {

            for (int i = left; i <= right; i++) result[index++] = matrix[upper][i];

        } else if (left == right && upper == lower) {

            result[index] = matrix[upper][left];

        }

        return result;

    }

}