顺时针打印二维数组(螺旋输出数组)
题目:给定一个数组,将该数组从第一个元素开始顺时针打印出来。
分析:先考虑打印周围一圈的问题,然后可以使用递归求解,直到最后全都打印完为止。也就是先打印最外围的数,然后对于元二维数组中间的数组作为一个新的数组,重新打印该新数组的外围的数,直到最后全部打印完为止。 代码如下:
package problem2; /** * @author Hutongling */ public class 顺时针打印二维数组 { static void print2DArray(int data[][]) { if (data[0].length == 1) { for (int i = 0; i < data.length; i++) System.out.print(data[i][0] + " "); return; } else if (data.length == 1 && data[0].length != 1) { for (int i = 0; i < data[0].length; i++) System.out.print(data[0][i] + " "); return; } else if (data.length > 1) { int row = data.length; int column = data[0].length; int i = 0; for (i = 0; i < column; i++) System.out.print(data[0][i] + " "); for (i = 1; i < row; i++) System.out.print(data[i][column - 1] + " "); for (i = column - 2; i >= 0; i--) System.out.print(data[row - 1][i] + " "); for (i = row - 2; i >= 1; i--) System.out.print(data[i][0] + " "); if (row - 2 > 0 && column - 2 > 0) { int subData[][] = new int[row - 2][column - 2]; for (i = 1; i < row - 1; i++) for (int j = 1; j < column - 1; j++) subData[i - 1][j - 1] = data[i][j]; data = null; print2DArray(subData); } } } public static void main(String[] args) { int data0[][] = { { 1, 2, 3, 4, 5, 6 }}; int data[][] = { { 1, 2, 3, 4, 5, 6 }, { 2, 3, 4, 5, 6, 7 }, { 3, 4, 5, 6, 7, 8 }, { 4, 5, 6, 7, 8, 9 } }; int data1[][] = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 16 } }; int data2[][] = { { 1, 2, 3, 4, 5 }, { 2, 3, 4, 5, 6 }, { 3, 4, 5, 6, 7 }, { 4, 5, 6, 7, 8 }, { 5, 6, 7, 8, 9 } }; int data3[][]={ { 1},{ 2},{ 3},{ 4}}; print2DArray(data0); System.out.println(); print2DArray(data); System.out.println(); print2DArray(data1); System.out.println(); print2DArray(data2); System.out.println(); print2DArray(data3); } }
结果如下: 1 2 3 4 5 6 1 2 3 4 5 6 7 8 9 8 7 6 5 4 3 2 3 4 5 6 7 6 5 4 1 2 3 4 8 12 16 15 14 13 9 5 6 7 11 10 1 2 3 4 5 6 7 8 9 8 7 6 5 4 3 2 3 4 5 6 7 6 5 4 5 1 2 3 4