获取列表元素的组合列表

时间:2014-01-30 12:23:22

标签: java combinatorics

假设我有3个列表:['q','w'],['a','s'],['z','x']。如何从这些列表中获取可能的组合列表?所以我得到一个列表[['q','a','z'],['q','s','z']]等等。我为两个方法制作了一个方法,但是无法为N个列表找到一个方法:

static <E> ArrayList combine(ArrayList<E> one,ArrayList<E> two)
{
    ArrayList<ArrayList<E>> combs=new ArrayList<ArrayList<E>>();
    for(E e:one)
    {
        for(E e2:two)
        {
            ArrayList ps=new ArrayList();
            ps.add(e);
            ps.add(e2);
            combs.add(ps);
        }
    }
    return combs;
}

我发现这是由Guava的Sets.cartesianProduct完成的。​​

4 个答案:

答案 0 :(得分:5)

懒惰(使用番石榴):

Set<List<String>> result = Sets.cartesianProduct(
                ImmutableSet.of("q", "w"),
                ImmutableSet.of("a", "s"),
                ImmutableSet.of("z", "x")
        );

System.out.println(result);

输出:

[ [q, a, z], [q, a, x], [q, s, z], [q, s, x], [w, a, z], [w, a, x], [w, s, z], [w, s, x] ]

答案 1 :(得分:2)

你需要N个嵌套循环,这就是它的难点。

您可以使用递归来实现此目的。

static <E> ArrayList combine(ArrayList<E> soFar, ArrayList<E>... lists)
{
    // Rather than constantly making and remaking this list could just use one
    // and pass it around and add stuff to it. This works though.
    ArrayList<ArrayList<E>> combs=new ArrayList<ArrayList<E>>();

    // Loop through the first list looking for elements
    for(E e:lists[0])
    {
       // Create a new List to build this combination
       ArrayList<E> temp = new ArrayList<>(soFar);
       // Add this element to the combination
       temp.add(e);
       // If there are more lists recurse down
       if (lists.length > 1) {
           // Use recursion to add all combinations of the remaining lists
           combs.addAll(combine(temp, lists.subList(1)));
       } else {
           // There are no more lists so we are done, add temp to the combos
           combs.add(temp);
       }
    }
    return combs;
}


// Call this method to start things going, the other one should probably be private
static <E> ArrayList combine(ArrayList<E>... lists)
    return combine(new ArrayList<E>(), lists);
}

答案 2 :(得分:1)

您可能希望查看课程https://github.com/javagl/Combinatorics/blob/master/src/main/java/de/javagl/utils/math/combinatorics/MixedRangeCombinationIterable.java(“独立”课程,只需复制并插入项目中)

使用示例:

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

public class Combinations
{
    public static void main(String[] args)
    {
        List<List<Character>> lists = new ArrayList<List<Character>>();
        lists.add(Arrays.asList('q','w'));
        lists.add(Arrays.asList('a','s'));
        lists.add(Arrays.asList('z','x'));

        MixedRangeCombinationIterable<Character> iterable = 
            new MixedRangeCombinationIterable<Character>(lists);
        for (List<Character> element : iterable)
        {
            System.out.println(element);
        }
    }
}

您实际上是在计算输入集{/ 3}}的元素

答案 3 :(得分:0)

up你可以使用构造函数(Parameters ...)创建一个内部类,这样就可以放置一个处理所有组合的类的列表