Question

我试图在没有重复的情况下找到数组{"A","B","C"}中的字符串组合，并且应该在子集中保留元素的顺序。所需订单为[["B","C"], ["A","C"], ["A","B"], ["A","B","C"], ["A"], ["C"], ["B"]]。我尝试使用此question中的答案编写逻辑，并发现元素的顺序不会被保留。

public static Set <JSONArray> getCombinations( int k , JSONArray properties )
        {
            Set <JSONArray> combinations = new LinkedHashSet <JSONArray>();
            try
                {
                    if ( k == 0 )
                        {
                            combinations.add( new JSONArray() );
                            return combinations;
                        }
                    for ( int i = 0 ; i < properties.length() ; i++ )
                        {
                            String element = properties.getString( i );
                            JSONArray sublist = getSublist( properties , i + 1 );
                            combinations.add( sublist );
                            Set <JSONArray> combinations2 = getCombinations( k - 1 , sublist );
                            for ( JSONArray previous : combinations2 )
                                {

                                    previous.put( element );
                                    combinations.add( previous );
                                }
                        }
                }
            catch ( Exception e )
                {
                    System.out.println( "Exception :: " + e );
                }
            return combinations;
        }

    public static JSONArray getSublist( JSONArray list , int i ) throws JSONException
        {
            JSONArray sublist = new JSONArray();
            for ( int j = i ; j < list.length() ; j++ )
                {
                    sublist.put( list.getString( j ) );
                }
            return reverseArray( sublist );
        }

输出为:: [["B","C"], ["C","A"], ["B","A"], ["C","B","A"], ["A"], ["C"], ["B"]]。但我需要保存的命令，如[“C”，“A”]应该是[“A”，“C”]。任何想法都会有所帮助。

PS：子集的顺序无关紧要，但子集内元素的顺序为。

Answer 1

组合可以用数字表示 - 以二进制形式，每个位置的数字表示元素是否存在。例如。 5 = 101 - ＆gt; {A，C}

因此，让我们迭代组合=数字范围＆lt; 0..2 ^ n-1＆gt;并获得与数字对应的元素，这意味着那些索引存在于组合的二进制表示中。

 public class Combins {

            static String[] a = new String[] { "A", "B", "C" };

            public static void main(final String[] args) {

                final int maxbit = 1 << a.length;

                //for each combination given by a (binary) number 'p'...
                for (int p = 0; p < maxbit; p++) {
                    final List<String> res = new ArrayList<String>();

                    //evaluate if array 'a' element at index 'i' is present in combination (and include it if so)
                    for (int i = 0; i < a.length; i++) {
                        if ((1 << i & p) > 0) {
                            res.add(a[i]);
                        }
                    }
                    System.out.println(Arrays.toString(res.toArray()));
                }
            }
        }

输出是：

[]
[A]
[B]
[A, B]
[C]
[A, C]
[B, C]
[A, B, C]

在Java中查找数组中所有字符串的组合

1 个答案: