查找数组中长度为 k 的所有子集 [英] Find all subsets of length k in an array

问题描述

给定一组 {1,2,3,4,5...n} 的 n 个元素，我们需要找到长度为 k 的所有子集.

Given a set {1,2,3,4,5...n} of n elements, we need to find all subsets of length k .

例如，如果 n = 4 且 k = 2，output 将是 {1, 2}, {1, 3}, {1, 4}, {2,3}, {2, 4}, {3, 4}.

For example, if n = 4 and k = 2, the output would be {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}.

我什至不知道如何开始.我们不必使用 next_permutation 等内置库函数.

I am not even able to figure out how to start. We don't have to use the inbuilt library functions like next_permutation etc.

需要 C/C++ 或 Java 中的算法和实现.

Need the algorithm and implementation in either C/C++ or Java.

推荐答案

递归是您完成此任务的朋友.

Recursion is your friend for this task.

对于每个元素 - 猜测"它是否在当前子集中，并递归调用猜测和您可以从中选择的较小超集.对是"和否"猜测都这样做 - 将产生所有可能的子集.
在停止子句中可以轻松地将自己限制在一定的长度.

For each element - "guess" if it is in the current subset, and recursively invoke with the guess and a smaller superset you can select from. Doing so for both the "yes" and "no" guesses - will result in all possible subsets.
Restraining yourself to a certain length can be easily done in a stop clause.

Java 代码:

private static void getSubsets(List<Integer> superSet, int k, int idx, Set<Integer> current,List<Set<Integer>> solution) {
    //successful stop clause
    if (current.size() == k) {
        solution.add(new HashSet<>(current));
        return;
    }
    //unseccessful stop clause
    if (idx == superSet.size()) return;
    Integer x = superSet.get(idx);
    current.add(x);
    //"guess" x is in the subset
    getSubsets(superSet, k, idx+1, current, solution);
    current.remove(x);
    //"guess" x is not in the subset
    getSubsets(superSet, k, idx+1, current, solution);
}

public static List<Set<Integer>> getSubsets(List<Integer> superSet, int k) {
    List<Set<Integer>> res = new ArrayList<>();
    getSubsets(superSet, k, 0, new HashSet<Integer>(), res);
    return res;
}

调用方式:

List<Integer> superSet = new ArrayList<>();
superSet.add(1);
superSet.add(2);
superSet.add(3);
superSet.add(4);
System.out.println(getSubsets(superSet,2));

将产生:

[[1, 2], [1, 3], [1, 4], [2, 3], [2, 4], [3, 4]]

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