Question

我有一系列元素（偶数），我想在夫妻中获得所有可能的组合。

如果数组是

String[] test = new String[4]; 
String[] test= {"e1","e2","e3","e4"};

输出应为：

Combination1 e1-e2 , e3-e4
Combination2 e1-e3 , e2-e4
Combination3 e1-e4 , e2-e3
Combination4 e4-e1 , e3-e2
Combination5 e3-e1 , e4-e2
Combination6 e2-e1 , e4-e3

Answer 1

写两个for循环：

for (int i =0; i < test.length; i++) {
    for (int j = i + 1; j < test.length; j++) {
       // code to print a[i] - a[j]
    }
 }

Answer 2

Link to previous answer

String[] test = {"e1","e2","e3","e4"};

for (int i = 0; i < test.length; i++) {

    for (int j = i + 1; j < test.length; j++) {

        System.out.print(test[i] + " - " + test[j]);

        boolean foundExtra = false;

        for (int k = 0; k < test.length && !foundExtra; k++)
        {
            if (k != j && k != i)
            {
                for (int l = 0; l < test.length; l++)
                {
                    if (l != k && l != j && l != i)
                    {
                        System.out.println(" , " + test[k] + " - " + test[l]);
                        foundExtra = true;
                        break;
                    }
                }
            }
        }
    }
}

会给出输出：

e1 - e2 , e3 - e4
e1 - e3 , e2 - e4
e1 - e4 , e2 - e3
e2 - e3 , e1 - e4
e2 - e4 , e1 - e3
e3 - e4 , e1 - e2

这不是你在问题中输出的输出，但我相信这是你想要的输出，从运动队评论来判断。

不要害怕循环 - 我一次尝试这样做是因为我觉得很容易丢掉它。

我在想什么：找到所有组合，就像我之前的回答一样。这就是我的开始。我做的下一件事（4个循环中的最后2个 - k和l）检查了剩下的其他团队。

因此，循环浏览所有元素，检查i和j是否尚未使用，然后是k。然后，再次遍历所有元素，检查i，j和k是否尚未使用，那是l。

Answer 3

这适用于任意偶数个元素的数组（并且是通用的，因此除了字符串之外还应该用于其他内容）：

import java.lang.reflect.Array;
import java.util.ArrayList;
import java.util.LinkedList;

public class Q2112634 {

    public static class Pair<X> {
        final X first;
        final X second;

        public Pair( final X one, final X two){
            first = one;
            second = two;
        }

        public String toString(){
            return first.toString() + "-" + second.toString(); 
        }

        public String toReverseString(){
            return second.toString() + "-" + first.toString(); 
        }
    }

    @SuppressWarnings("unchecked")
    public static <X> void getCombinations( final ArrayList<Pair<X>[]> combinations, final LinkedList<Pair<X>> pairs, final LinkedList<X> list ) {
//      System.out.println(list.size());
        final X first = list.removeFirst();
        for ( int i = 0; i < list.size(); ++i ) {
            final X second = list.remove( i );
            final Pair<X> p = new Pair<X>( first, second );
            pairs.addLast( p );
            if ( list.size() == 0 )
            {
                combinations.add( pairs.toArray( (Pair[]) Array.newInstance( p.getClass(), pairs.size() ) ) );
            }
            else
            {
                getCombinations( combinations, pairs, list );
            }
            list.add( i, second );
            pairs.removeLast();
        }
        list.addFirst( first );
    }

    static <E> String arrayToString( E[] arr ) {
        if ( arr.length == 0 )
            return "";
        final StringBuffer str = new StringBuffer();
        str.append( arr[0].toString() );
        for ( int i = 1; i < arr.length; ++i )
        {
            str.append( ',' );
            str.append( arr[i].toString() );
        }
        return str.toString();
    }

    public static void main(String[] args) {
        String[] test = { "e1", "e2", "e3", "e4" };

        int num_combinations = 1;
        for ( int i = test.length - 1; i > 1; i = i - 2 )
            num_combinations *= i;

        final ArrayList<Pair<String>[]> combinations = new ArrayList<Pair<String>[]>( num_combinations );
        final LinkedList<String> strings = new LinkedList<String>();
        for ( String s : test )
            strings.add( s );

        getCombinations( combinations, new LinkedList<Pair<String>>(), strings );

        System.out.println( "-----Combinations-----" );
        int i = 1;
        for ( Pair<String>[] combination: combinations ){
            System.out.println( "Combination " + (i++) + " " + arrayToString( combination ) );
        }

        System.out.println( "-----Permutations-----" );
        i = 1;
        for ( Pair<String>[] combination: combinations ){
            for ( int j = 0; j < Math.pow( 2, combination.length ); ++j )
            {
                System.out.print( "Permutation " + (i++) + " " );
                for ( int k = 0; k < combination.length; ++k )
                {
                    if ( k > 0 )
                        System.out.print(',');
                    if ( (j & ( (int) Math.pow( 2, k ) ) ) == 0 )
                        System.out.print( combination[k].toString() );
                    else
                        System.out.print( combination[k].toReverseString() );
                }
                System.out.println();
            }
        }
    }
}

<强>输出

-----Combinations-----
Combination 1 e1-e2,e3-e4
Combination 2 e1-e3,e2-e4
Combination 3 e1-e4,e2-e3
-----Permutations-----
Permutation 1 e1-e2,e3-e4
Permutation 2 e2-e1,e3-e4
Permutation 3 e1-e2,e4-e3
Permutation 4 e2-e1,e4-e3
Permutation 5 e1-e3,e2-e4
Permutation 6 e3-e1,e2-e4
Permutation 7 e1-e3,e4-e2
Permutation 8 e3-e1,e4-e2
Permutation 9 e1-e4,e2-e3
Permutation 10 e4-e1,e2-e3
Permutation 11 e1-e4,e3-e2
Permutation 12 e4-e1,e3-e2

Answer 4

试试这个：

String[] test= {"e1","e2","e3","e4"};
int count=1;
for(int i=0; i<test.length/2; i++)
    for(int k=0, j=0; k<test.length; k++,j++) {
        if(i==k) k++; if(j==i+test.length/2) j++;
        System.out.print("Permutatation "+count+": ");
        System.out.println(test[i]+"-"+test[k]+", "+test[i+test.length/2]+"-"+test[j]);
        count++;
    }

输出：

Permutatation 1: e1-e2, e3-e1
Permutatation 2: e1-e3, e3-e2
Permutatation 3: e1-e4, e3-e4
Permutatation 4: e2-e1, e4-e1
Permutatation 5: e2-e3, e4-e2
Permutatation 6: e2-e4, e4-e3

Java：成对的数组元素组合

4 个答案: