小编给大家分享一下Java如何实现笛卡尔积算法，相信大部分人都还不怎么了解，因此分享这篇文章给大家参考一下，希望大家阅读完这篇文章后大有收获，下面让我们一起去了解一下吧！
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具体如下：
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
最近碰到了一个笛卡尔积的算法要求，比如传递过来的参数是"1,3,6,7==4,5,8,9==3,4==43,45,8,9==35,4"，则返回的是一个list，如[1,4,3,43,35][1,4,3,43,4][1,4,3,45,35]……，该list包含是4*4*2*4*2=256个元素，现在的思路是这样的：
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
public class DescartesTest {
/**
* 获取N个集合的笛卡尔积
*
* 说明：假如传入的字符串为："1,2,3==5,6==7,8"
* 转换成字符串数组为：[[1, 2, 3], [5, 6], [7, 8]]
* a=[1, 2, 3]
* b=[5, 6]
* c=[7, 8]
* 其大小分别为：a_length=3，b_length=2，c_length=2，
* 目标list的总大小为：totalSize=3*2*2 = 12
* 对每个子集a，b，c，进行循环次数=总记录数/(元素个数*后续集合的笛卡尔积个数)
* 对a中的每个元素循环次数=总记录数/(元素个数*后续集合的笛卡尔积个数)=12/(3*4)=1次，每个元素每次循环打印次数：后续集合的笛卡尔积个数=2*2个
* 对b中的每个元素循环次数=总记录数/(元素个数*后续集合的笛卡尔积个数)=12/(2*2)=3次，每个元素每次循环打印次数：后续集合的笛卡尔积个数=2个
* 对c中的每个元素循环次数=总记录数/(元素个数*后续集合的笛卡尔积个数)=12/(2*1)=6次，每个元素每次循环打印次数：后续集合的笛卡尔积个数=1个
*
* 运行结果：
* [[1, 2, 3], [5, 6], [7, 8]]
1,5,7,
1,5,8,
1,6,7,
1,6,8,
2,5,7,
2,5,8,
2,6,7,
2,6,8,
3,5,7,
3,5,8,
3,6,7,
3,6,8]
从结果中可以看到：
a中的每个元素每个元素循环1次，每次打印4个
b中的每个元素每个元素循环3次，每次打印2个
c中的每个元素每个元素循环6次，每次打印1个
*
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
String str ="1,3,6,7==4,5,8,9==3,4==43,45,8,9==35,4";
List result = descartes(str);
System.out.println(result);
}
@SuppressWarnings("rawtypes")
public static List descartes(String str) {
String[] list = str.split("==");
List strs = new ArrayList();
for(int i=0;i运行结果输出：
[[1, 3, 6, 7], [4, 5, 8, 9], [3, 4], [43, 45, 8, 9], [35, 4]]
[1,4,3,43,35, 1,4,3,43,4, 1,4,3,45,35, 1,4,3,45,4, 1,4,3,8,35, 1,4,3,8,4, 1,4,3,9,35, 1,4,3,9,4, 1,4,4,43,35, 1,4,4,43,4, 1,4,4,45,35, 1,4,4,45,4, 1,4,4,8,35, 1,4,4,8,4, 1,4,4,9,35, 1,4,4,9,4, 1,5,3,43,35, 1,5,3,43,4, 1,5,3,45,35, 1,5,3,45,4, 1,5,3,8,35, 1,5,3,8,4, 1,5,3,9,35, 1,5,3,9,4, 1,5,4,43,35, 1,5,4,43,4, 1,5,4,45,35, 1,5,4,45,4, 1,5,4,8,35, 1,5,4,8,4, 1,5,4,9,35, 1,5,4,9,4, 1,8,3,43,35, 1,8,3,43,4, 1,8,3,45,35, 1,8,3,45,4, 1,8,3,8,35, 1,8,3,8,4, 1,8,3,9,35, 1,8,3,9,4, 1,8,4,43,35, 1,8,4,43,4, 1,8,4,45,35, 1,8,4,45,4, 1,8,4,8,35, 1,8,4,8,4, 1,8,4,9,35, 1,8,4,9,4, 1,9,3,43,35, 1,9,3,43,4, 1,9,3,45,35, 1,9,3,45,4, 1,9,3,8,35, 1,9,3,8,4, 1,9,3,9,35, 1,9,3,9,4, 1,9,4,43,35, 1,9,4,43,4, 1,9,4,45,35, 1,9,4,45,4, 1,9,4,8,35, 1,9,4,8,4, 1,9,4,9,35, 1,9,4,9,4, 3,4,3,43,35, 3,4,3,43,4, 3,4,3,45,35, 3,4,3,45,4, 3,4,3,8,35, 3,4,3,8,4, 3,4,3,9,35, 3,4,3,9,4, 3,4,4,43,35, 3,4,4,43,4, 3,4,4,45,35, 3,4,4,45,4, 3,4,4,8,35, 3,4,4,8,4, 3,4,4,9,35, 3,4,4,9,4, 3,5,3,43,35, 3,5,3,43,4, 3,5,3,45,35, 3,5,3,45,4, 3,5,3,8,35, 3,5,3,8,4, 3,5,3,9,35, 3,5,3,9,4, 3,5,4,43,35, 3,5,4,43,4, 3,5,4,45,35, 3,5,4,45,4, 3,5,4,8,35, 3,5,4,8,4, 3,5,4,9,35, 3,5,4,9,4, 3,8,3,43,35, 3,8,3,43,4, 3,8,3,45,35, 3,8,3,45,4, 3,8,3,8,35, 3,8,3,8,4, 3,8,3,9,35, 3,8,3,9,4, 3,8,4,43,35, 3,8,4,43,4, 3,8,4,45,35, 3,8,4,45,4, 3,8,4,8,35, 3,8,4,8,4, 3,8,4,9,35, 3,8,4,9,4, 3,9,3,43,35, 3,9,3,43,4, 3,9,3,45,35, 3,9,3,45,4, 3,9,3,8,35, 3,9,3,8,4, 3,9,3,9,35, 3,9,3,9,4, 3,9,4,43,35, 3,9,4,43,4, 3,9,4,45,35, 3,9,4,45,4, 3,9,4,8,35, 3,9,4,8,4, 3,9,4,9,35, 3,9,4,9,4, 6,4,3,43,35, 6,4,3,43,4, 6,4,3,45,35, 6,4,3,45,4, 6,4,3,8,35, 6,4,3,8,4, 6,4,3,9,35, 6,4,3,9,4, 6,4,4,43,35, 6,4,4,43,4, 6,4,4,45,35, 6,4,4,45,4, 6,4,4,8,35, 6,4,4,8,4, 6,4,4,9,35, 6,4,4,9,4, 6,5,3,43,35, 6,5,3,43,4, 6,5,3,45,35, 6,5,3,45,4, 6,5,3,8,35, 6,5,3,8,4, 6,5,3,9,35, 6,5,3,9,4, 6,5,4,43,35, 6,5,4,43,4, 6,5,4,45,35, 6,5,4,45,4, 6,5,4,8,35, 6,5,4,8,4, 6,5,4,9,35, 6,5,4,9,4, 6,8,3,43,35, 6,8,3,43,4, 6,8,3,45,35, 6,8,3,45,4, 6,8,3,8,35, 6,8,3,8,4, 6,8,3,9,35, 6,8,3,9,4, 6,8,4,43,35, 6,8,4,43,4, 6,8,4,45,35, 6,8,4,45,4, 6,8,4,8,35, 6,8,4,8,4, 6,8,4,9,35, 6,8,4,9,4, 6,9,3,43,35, 6,9,3,43,4, 6,9,3,45,35, 6,9,3,45,4, 6,9,3,8,35, 6,9,3,8,4, 6,9,3,9,35, 6,9,3,9,4, 6,9,4,43,35, 6,9,4,43,4, 6,9,4,45,35, 6,9,4,45,4, 6,9,4,8,35, 6,9,4,8,4, 6,9,4,9,35, 6,9,4,9,4, 7,4,3,43,35, 7,4,3,43,4, 7,4,3,45,35, 7,4,3,45,4, 7,4,3,8,35, 7,4,3,8,4, 7,4,3,9,35, 7,4,3,9,4, 7,4,4,43,35, 7,4,4,43,4, 7,4,4,45,35, 7,4,4,45,4, 7,4,4,8,35, 7,4,4,8,4, 7,4,4,9,35, 7,4,4,9,4, 7,5,3,43,35, 7,5,3,43,4, 7,5,3,45,35, 7,5,3,45,4, 7,5,3,8,35, 7,5,3,8,4, 7,5,3,9,35, 7,5,3,9,4, 7,5,4,43,35, 7,5,4,43,4, 7,5,4,45,35, 7,5,4,45,4, 7,5,4,8,35, 7,5,4,8,4, 7,5,4,9,35, 7,5,4,9,4, 7,8,3,43,35, 7,8,3,43,4, 7,8,3,45,35, 7,8,3,45,4, 7,8,3,8,35, 7,8,3,8,4, 7,8,3,9,35, 7,8,3,9,4, 7,8,4,43,35, 7,8,4,43,4, 7,8,4,45,35, 7,8,4,45,4, 7,8,4,8,35, 7,8,4,8,4, 7,8,4,9,35, 7,8,4,9,4, 7,9,3,43,35, 7,9,3,43,4, 7,9,3,45,35, 7,9,3,45,4, 7,9,3,8,35, 7,9,3,8,4, 7,9,3,9,35, 7,9,3,9,4, 7,9,4,43,35, 7,9,4,43,4, 7,9,4,45,35, 7,9,4,45,4, 7,9,4,8,35, 7,9,4,8,4, 7,9,4,9,35, 7,9,4,9,4]
递归算法：
public static void fn(List list,String[] arr,String str){
//迭代list
List li = new ArrayList();
for(int i=0;i以上是“Java如何实现笛卡尔积算法”这篇文章的所有内容，感谢各位的阅读！相信大家都有了一定的了解，希望分享的内容对大家有所帮助，如果还想学习更多知识，欢迎关注创新互联行业资讯频道！
分享题目：Java如何实现笛卡尔积算法
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