如何用Java实现排列组合算法

private static Set<Set<String>> exhaustion() {
    List<String> m = Arrays.asList("a", "b", "c", "d", "e");
    Set<Set<String>> result = new HashSet<>();
    int count = 3;
    for (int a = 1; a < m.size(); a++) {
        for (int b = 0; b < m.size(); b++) {
            for (int c = 0; c < m.size(); c++) {
                Set<String> tempCollection = new HashSet<>();
                tempCollection.add(m.get(a));
                tempCollection.add(m.get(b));
                tempCollection.add(m.get(c));
                // 如果三个元素中有重复的会被 Set 排重，导致 Set 的大小不为 3
                if (tempCollection.size() == count) {
                    result.add(tempCollection);
                }
            }
        }
    }

    return result;
}

public class Combination {

    public static void main(String[] args) {
        List<String> m = Arrays.asList("a", "b", "c", "d", "e");
        int n = 5;

        Set<Set<String>> combinationAll = new HashSet<>();
        // 先将问题分解成 五取一、五取二... 等的全排列
        for (int c = 1; c <= n; c++) {
            combinationAll.addAll(combination(m, new ArrayList<>(), c));
        }

        System.out.println(combinationAll);
    }

    private static Set<Set<String>> combination(List<String> remainEle, List<String> tempCollection, int fetchCount) {
        if (fetchCount == 1) {
            Set<Set<String>> eligibleCollections = new HashSet<>();
            // 在只差一个元素的情况下，遍历剩余元素为每个临时集合生成多个满足条件的集合
            for (String ele : remainEle) {
                Set<String> collection = new HashSet<>(tempCollection);
                collection.add(ele);
                eligibleCollections.add(collection);
            }
            return eligibleCollections;
        }

        fetchCount--;
        Set<Set<String>> result = new HashSet<>();
        // 差多个元素时，从剩余元素中取出一个，产生多个临时集合，还需要取 count-- 个元素。
        for (int i = 0; i < remainEle.size(); i++) {
            List<String> collection = new ArrayList<>(tempCollection);
            List<String> tempRemain = new ArrayList<>(remainEle);
            collection.add(tempRemain.remove(i));
            result.addAll(combination(tempRemain, collection, fetchCount));
        }
        return result;
    }
}

public class Combination {

    public static void main(String[] args) {
        String[] m = {"A", "B", "C", "D", "E"};
        Set<Set<String>> combinationAll = combination(m);
        System.out.println(combinationAll);

    }

    private static Set<Set<String>> combination(String[] m) {
        Set<Set<String>> result = new HashSet<>();

        for (int i = 1; i < Math.pow(2, m.length) - 1; i++) {
            Set<String> eligibleCollections = new HashSet<>();
            // 依次将数字 i 与 2^n 按位与，判断第 n 位是否为 1
            for (int j = 0; j < m.length; j++) {
                if ((i & (int) Math.pow(2, j)) == Math.pow(2, j)) {
                    eligibleCollections.add(m[j]);
                }
            }
            result.add(eligibleCollections);
        }
        return result;
    }
}