笛卡尔积算法的Java实现

（1）循环内，每次只有一列向下移一个单元格，就是CounterIndex指向的那列。

（2）如果该列到尾部了，则这列index重置为0，而CounterIndex则指向前一列，相当于进位，把前列的index加一。

（3）最后，由生成的行数来控制退出循环。

public class Test {

private static String[] aa = { "aa1", "aa2" };
private static String[] bb = { "bb1", "bb2", "bb3" };
private static String[] cc = { "cc1", "cc2", "cc3", "cc4" };
private static String[][] xyz = { aa, bb, cc };
private static int counterIndex = xyz.length - 1;
private static int[] counter = { 0, 0, 0 };

public static void main(String[] args) throws Exception {

for (int i = 0; i < aa.length * bb.length * cc.length; i++) {
System.out.print(aa[counter[0]]);
System.out.print("\t");
System.out.print(bb[counter[1]]);
System.out.print("\t");
System.out.print(cc[counter[2]]);
System.out.println();

handle();
}
}

public static void handle() {
counter[counterIndex]++;
if (counter[counterIndex] >= xyz[counterIndex].length) {
counter[counterIndex] = 0;
counterIndex--;
if (counterIndex >= 0) {
handle();
}
counterIndex = xyz.length - 1;
}
}

}

输出共2*3*4=24行：

aa1 bb1 cc1

aa1 bb1 cc2

aa1 bb1 cc3

aa1 bb1 cc4

aa1 bb2 cc1

aa1 bb2 cc2

aa1 bb2 cc3

aa1 bb2 cc4

aa1 bb3 cc1

aa1 bb3 cc2

aa1 bb3 cc3

aa1 bb3 cc4

aa2 bb1 cc1

aa2 bb1 cc2

aa2 bb1 cc3

aa2 bb1 cc4

aa2 bb2 cc1

aa2 bb2 cc2

aa2 bb2 cc3

aa2 bb2 cc4

aa2 bb3 cc1

aa2 bb3 cc2

aa2 bb3 cc3

aa2 bb3 cc4