使用递归排列数字的 ArrayList

Question

我正在尝试通过创建 ArrayList 的排列来学习递归：

 {1,2,3}

但递归调用的概念一直在我脑海中浮现。我知道如何做一个迭代解决方案。但是有没有一种系统的方法可以将我的迭代解决方案转换为递归解决方案？

private static ArrayList<ArrayList<Integer>> permutate(ArrayList<Integer> list) {
    ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();

    result.add(new ArrayList<Integer>());

    for (int i = 0; i < list.size(); i++) {
        ArrayList<ArrayList<Integer>> current = new ArrayList<ArrayList<Integer>>();

        for (ArrayList<Integer> l : result) {
            for (int j = 0; j < l.size()+1; j++) {
                l.add(j, list.get(i));

                ArrayList<Integer> temp = new ArrayList<Integer>(l);
                current.add(temp);

                l.remove(j);
            }
        }

        result = new ArrayList<ArrayList<Integer>>(current);
    }

    return result;

}

Answer 1

你来了，伙计：

main:
ArrayList<Integer> list = new ArrayList<Integer>();
    list.add(1);
    list.add(2);
    list.add(3);
    ArrayList<ArrayList<Integer>> ans = permutate(list, null, 0);


private static ArrayList<ArrayList<Integer>> permutate(
        ArrayList<Integer> list, ArrayList<Integer> curlist, int cur) {
    if (cur == 0) {
        ArrayList<ArrayList<Integer>> totalAns = new ArrayList<ArrayList<Integer>>();
        for (Iterator<Integer> iterator = list.iterator(); iterator
                .hasNext();) {
            Integer integer = (Integer) iterator.next();
            ArrayList<Integer> tmp3 = new ArrayList<Integer>();
            tmp3.add(integer);
            ArrayList<ArrayList<Integer>> firstAns = permutate(list, tmp3,
                    cur + 1);
            for (Iterator<ArrayList<Integer>> iterator2 = firstAns
                    .iterator(); iterator2.hasNext();) {
                ArrayList<Integer> arrayList = (ArrayList<Integer>) iterator2
                        .next();
                totalAns.add(arrayList);

            }
        }
        return totalAns;
    }
    if (cur == list.size()) {
        ArrayList<ArrayList<Integer>> tmp2 = new ArrayList<ArrayList<Integer>>();
        tmp2.add(curlist);
        return tmp2;
    }

    ArrayList<ArrayList<Integer>> ans = new ArrayList<ArrayList<Integer>>();
    for (int i = 0; i < list.size(); i++) {
        if (!curlist.contains(list.get(i))) {
            @SuppressWarnings("unchecked")
            ArrayList<Integer> tmp = (ArrayList<Integer>) curlist.clone();
            tmp.add(list.get(i));
            ArrayList<ArrayList<Integer>> recAns = permutate(list, tmp,
                    cur + 1);
            for (int k = 0; k < recAns.size(); k++) {
                ans.add(recAns.get(k));
            }
        }

    }
    return ans;
}

Answer 2

public static List<List<Integer>> listPermutations(List<Integer> list) {

    if (list.size() == 0) {
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        result.add(new ArrayList<Integer>());
        return result;
    }

    List<List<Integer>> returnMe = new ArrayList<List<Integer>>();

    Integer firstElement = list.remove(0);

    List<List<Integer>> recursiveReturn = listPermutations(list);
    for (List<Integer> li : recursiveReturn) {

        for (int index = 0; index <= li.size(); index++) {
            List<Integer> temp = new ArrayList<Integer>(li);
            temp.add(index, firstElement);
            returnMe.add(temp);
        }

    }
    return returnMe;
}

为了测试我使用了：

public static void main(String[] args) throws Exception {

    List<Integer> intList = new ArrayList<Integer>();
    intList.add(1);
    intList.add(2);
    intList.add(3);
    List<List<Integer>> myLists = listPermutations(intList);

    for (List<Integer> al : myLists) {
        String appender = "";
        for (Integer i : al) {
            System.out.print(appender + i);
            appender = " ";
        }
        System.out.println();
    }

}

这给了我输出：

1 2 3
2 1 3
2 3 1
1 3 2
3 1 2
3 2 1