我想找到一个整数集的子集。它是萨姆亚群的算法与回溯的第一步。我写了下面的code，但它并没有返回正确的答案：

I want to find the subsets of a set of integers. it is the first step of "Sum of Subsets" algorithm with backtracking. I have written the following code, but it doesn't return the correct answer:

code：

BTSum(0, nums);
///**************
ArrayList<Integer> list = new ArrayList<Integer>();

public static ArrayList<Integer> BTSum(int n, ArrayList<Integer> numbers) {
    if (n == numbers.size()) {
        for (Integer integer : list) {
            System.out.print(integer+", ");
        }
        System.out.println("********************");
        list.removeAll(list);
        System.out.println();
    } else {
        for (int i = n; i < numbers.size(); i++) {
            if (i == numbers.size() - 1) {
                list.add(numbers.get(i));
                BTSum(i + 1, numbers);
            } else {
                list.add(numbers.get(i));
                for (int j = i+1; j < numbers.size(); j++)
                BTSum(j, numbers);
            }
        }
    }

    return null;
}

例如，如果我想计算集合的子集= {1，3，5} 我的方法的结果是：

for example if I want to calculate the subsets of set = {1, 3, 5} The result of my method is :

 1, 3, 5, ********************

 5, ********************

 3, 5, ********************

 5, ********************

 3, 5, ********************

 5, ********************

我希望它产生

I want it to produce

1, 3, 5 
1, 5
3, 5
5

我认为这个问题是从零件 list.removeAll（名单）; 但我不知道如何纠正它。

I think the problem is from the part list.removeAll(list); but I dont know how to correct it.

推荐答案

你想要什么叫做 Powerset的 。下面是一个简单的实现它：

What you want is called a Powerset. Here is a simple implementation of it:

public static Set<Set<Integer>> powerSet(Set<Integer> originalSet) {
        Set<Set<Integer>> sets = new HashSet<Set<Integer>>();
        if (originalSet.isEmpty()) {
            sets.add(new HashSet<Integer>());
            return sets;
        }
        List<Integer> list = new ArrayList<Integer>(originalSet);
        Integer head = list.get(0);
        Set<Integer> rest = new HashSet<Integer>(list.subList(1, list.size()));
        for (Set<Integer> set : powerSet(rest)) {
            Set<Integer> newSet = new HashSet<Integer>();
            newSet.add(head);
            newSet.addAll(set);
            sets.add(newSet);
            sets.add(set);
        }
        return sets;
    }

我给你举个例子来解释如何算法工程的幂 {1，2，3} ：