Question

假设我有这个数组： [a，b，c，d] 我如何找到所有可能的组合，即 ab，abc，abcd .....等。 这需要包含重复项，因此 abcd 与 dcba 不同目的是找到所有组合并检查是否可以从不同的阵列进行相同的组合。我最初的尝试是：

for (int i = 0; i < values.length; i++) {
        String cur = values[i];

        for (int k = 0; k < values.length; k++) {
            if (i != k) {
                cur = cur.concat(values[k]);
                System.out.println(cur);
            }

        }
    }

给出了输出：

ab
abc
abcd
ba
bac
bacd
ca
cab
cabd
da
dab
dabc

显然不正确

这是针对我尝试改进的编程挑战，因此任何有关更快解决方案的建议都会有所帮助

Answer 1

这是你正在寻找的吗？

public static void main(String[] data) {
    ArrayList<Character> chars = new ArrayList<>(4);
    chars.add('a');
    chars.add('b');
    chars.add('c');
    chars.add('d');
    System.out.println(getPermutations("", chars));

}

private static ArrayList<String> getPermutations(String currentResult, ArrayList<Character> possibleChars) {
    ArrayList<String> result = new ArrayList<>(possibleChars.size());
    for (char append: possibleChars) {
        String permutation = currentResult + append; //create a new string with an additional character
        result.add(permutation); //add the permutation to the result
        if (possibleChars.size() > 0) {
            //make a new list with the appendable characters
            ArrayList<Character> possibleCharsUpdated = (ArrayList)possibleChars.clone();
            //from that list, exclude the character we just appended
            possibleCharsUpdated.remove(new Character(append));
            //merge the result of a recursive call of this method and the result we already had
            result.addAll(getPermutations(permutation, possibleCharsUpdated));
        }
    }
    return result;
}

在Java中查找数组中每个可能的元素组合

1 个答案: