求简化多重循环

[二酉子敬]

/**
* 测试调用的方法
*/
public static void testForFunction()
{
String[] betBunches = new String[]{"2", "3", "4"};
Object[][] list = new Object[][]{};
list[0] = new String[]{"周一001|胜|1", "周一001|平|2"};
list[1] = new String[]{"周二001|胜|3", "周二001|平|4"};
list[2] = new String[]{"周三001|胜|5"};
list[3] = new String[]{"周四001|胜|6"};
getBetList(betBunches, list);
}

/**
* 说 明： 计算投注注数
*
* @param betBunches 选择的串玩法
* @param list 投注场次和投注号码
* 创 建 人： zd
* 创建时间： 2018/12/25
**/
public static List<List<Object>> getBetList(String[] betBunches, Object[][] list)
{
Combination cb = new Combination(list);
List<List<Object>> cmNew = new ArrayList<>();
for (String betBunch : betBunches)
{
Integer num = Integer.valueOf(betBunch);
List<Object[][]> cm = cb.select(num);
for (int i = 0; i < cm.size(); i++)
{
Object[][] c = cm.get(i);
switch (num)
{
case 2:
for (Object c0 : c[0])
{
for (Object c1 : c[1])
{
List<Object> a2 = new ArrayList<>();
a2.add(c0);
a2.add(c1);
cmNew.add(a2);
}
}
break;
case 3:
for (Object c0 : c[0])
{
for (Object c1 : c[1])
{
for (Object c2 : c[2])
{
List<Object> a2 = new ArrayList<>();
a2.add(c0);
a2.add(c1);
a2.add(c2);
cmNew.add(a2);
}
}
}
break;
case 4:
for (Object c0 : c[0])
{
for (Object c1 : c[1])
{
for (Object c2 : c[2])
{
for (Object c3 : c[3])
{
List<Object> a3 = new ArrayList<>();
a3.add(c0);
a3.add(c1);
a3.add(c2);
a3.add(c3);
cmNew.add(a3);
}
}
}
}
break;
case 5:
for (Object c0 : c[0])
{
for (Object c1 : c[1])
{
for (Object c2 : c[2])
{
for (Object c3 : c[3])
{
for (Object c4 : c[4])
{
List<Object> a4 = new ArrayList<>();
a4.add(c0);
a4.add(c1);
a4.add(c2);
a4.add(c3);
a4.add(c4);
cmNew.add(a4);
}
}
}
}
}
break;
case 6:
for (Object c0 : c[0])
{
for (Object c1 : c[1])
{
for (Object c2 : c[2])
{
for (Object c3 : c[3])
{
for (Object c4 : c[4])
{
for (Object c5 : c[5])
{
List<Object> a5 = new ArrayList<>();
a5.add(c0);
a5.add(c1);
a5.add(c2);
a5.add(c3);
a5.add(c4);
a5.add(c5);
cmNew.add(a5);
}
}
}
}
}
}
break;
case 7:
for (Object c0 : c[0])
{
for (Object c1 : c[1])
{
for (Object c2 : c[2])
{
for (Object c3 : c[3])
{
for (Object c4 : c[4])
{
for (Object c5 : c[5])
{
for (Object c6 : c[6])
{
List<Object> a6 = new ArrayList<>();
a6.add(c0);
a6.add(c1);
a6.add(c2);
a6.add(c3);
a6.add(c4);
a6.add(c5);
a6.add(c6);
cmNew.add(a6);
}
}
}
}
}
}
}
break;
case 8:
for (Object c0 : c[0])
{
for (Object c1 : c[1])
{
for (Object c2 : c[2])
{
for (Object c3 : c[3])
{
for (Object c4 : c[4])
{
for (Object c5 : c[5])
{
for (Object c6 : c[6])
{
for (Object c7 : c[7])
{
List<Object> a7 = new ArrayList<>();
a7.add(c0);
a7.add(c1);
a7.add(c2);
a7.add(c3);
a7.add(c4);
a7.add(c5);
a7.add(c6);
a7.add(c7);
cmNew.add(a7);
}
}
}
}
}
}
}
}
break;
default:
}
}
}
Log.e("getBetList()", cmNew.toString());
return cmNew;
}

[十八道胡同]

这个业务是这样的，没啥好思路可以简化。。。

[二酉子敬]

引用 1 楼 LCL_data 的回复:

这个业务是这样的，没啥好思路可以简化。。。

are u 秀儿？

[咸哼酒家]

递归思想了解一波，
自己调用自己，结果当参数传递进去再得到结果
，最终return

[咸哼酒家]

case 1、2、3等，万一数字很大就很难办，
应当拿到记录条数，动态处理

[maradona1984]

没法测试,你自己测吧

    public static List<List<Object>> getBetList(String[] betBunches, Object[][] list) {
        Combination cb = new Combination(list);
        List<List<Object>> cmNew = new ArrayList<>();
        for (String betBunch : betBunches) {
            Integer num = Integer.valueOf(betBunch);
            List<Object[][]> cm = cb.select(num);
            for (int i = 0; i < cm.size(); i++) {
                Object[][] c = cm.get(i);
                List<Object> objectList = new ArrayList<>();
                combination(cmNew, objectList, c, 0);
            }
        }
        return cmNew;
    }
    
    public static void combination(List<List<Object>> cmNew, List<Object> objectList, Object[][] objectss,
        int index) {
        int nextIndex = index + 1;
        for (Object object : objectss[index]) {
            List<Object> tempList = new ArrayList<>(objectList);
            tempList.add(object);
            if (objectss.length > nextIndex) {
                combination(cmNew, tempList, objectss, nextIndex);
            } else {
                cmNew.add(tempList);
            }
        }
    }