Question

有人可以解释一下这段代码的工作原理吗？是否有可能以另一种方式写作？我只用ArrayList尝试过，但无法理解。

public static Set<Set<Integer>> combinations(List<Integer> groupSize, int k) {
    Set<Set<Integer>> allCombos = new HashSet<Set<Integer>> ();

    // base cases for recursion
    if (k == 0) {
        // There is only one combination of size 0, the empty team.
        allCombos.add(new HashSet<Integer>());
        return allCombos;
    }
    if (k > groupSize.size()) {
        // There can be no teams with size larger than the group size,
        // so return allCombos without putting any teams in it.
        return allCombos;
    }

    // Create a copy of the group with one item removed.
    List<Integer> groupWithoutX = new ArrayList<Integer> (groupSize);
    Integer x = groupWithoutX.remove(groupWithoutX.size()-1);

    Set<Set<Integer>> combosWithoutX = combinations(groupWithoutX, k);
    Set<Set<Integer>> combosWithX = combinations(groupWithoutX, k-1);
    for (Set<Integer> combo : combosWithX) {
        combo.add(x);
    }
    allCombos.addAll(combosWithoutX);
    allCombos.addAll(combosWithX);
    return allCombos;
}

Answer 1

The code returns a set of integer sets with all possible combinations of the integers in the first given set with k integers. I'll just demonstrate what it does with the case with list = [1,2,3] and k = 2.

groupWithoutX = [1,2].
x = 1.
k = 2.

Then, in the call combinations(groupWithoutX, k), the second base case is true, so the method just returns [1,2].


In the other recursive call, the recursion happens again.

groupWithoutX = [1].
x = 2.
k = 1.

The first call returns 1.
The second call returns an empty set.
However, x must be added to this empty set, so it becomes 2.

The second recursive call of the initial call of the method, then, returns a set of the sets of integers: [1] [2]

Back at the first call of the method, x = 3 must be added to each of the sets the second recursive call. [1] becomes [1,3] and [2] becomes [2,3].

The method returns all three created sets: [1,2] [1,3] [2,3].

In other words, this method is similar to the combination (nCr) in math, except it returns the sets created rather than just the number of sets.

递归方法返回一组所有组合

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