我试图生成所有可能的项目唯一组合.
I trying to generate all possible unique combination of items.
Ex: item1, item2, item3 Combinations: item1+item2+item3 item1+item2 item1+item3 item2+item3 item1 item2 item3我不知道如何解决这个问题?
I am unable to get an idea on how to solve this?
for(int i=0;i<size;i++){ for(int j=i+1;j<size;j++){ System.out.println(list.item(i)+list.item(j)); } }上面的代码当然适用于两个元素的所有唯一组合.但是不适用于3个元素对或更多..
The above code certainly works for all unique combination of two elements. But not for 3 element pair and more..
推荐答案如果您有N个项目,则计数从1到2 ^ N-1.每个数字都表示一个组合,如下所示:如果设置了位0(最低有效位),则item1在组合中.如果设置了位1,则item2处于组合中,依此类推.
If you have N items, count from 1 to 2^N-1. Each number represents a combination, like so: if bit 0 (the least significant bit) is set, item1 is in the combination. If bit 1 is set, item2 is in the combination, and so on.
如果您不想要1个项目的组合,请从3开始计数,并忽略所有2的幂的组合(4、8、16等).
If you don't want 1-item combinations, start counting at 3, and ignore all the combinations that are a power of 2 (4, 8, 16, etc...).
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生成项目的所有唯一组合
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