Question

假设我有几个List<T>，我会把它们放到另一个列表（或其他集合）中，所以我不知道有多少列表，直到我调用List＆gt; .size（）

以下列表为例：

list1=[1,2]  
list2=[3,4]
list3=[5,6]
....
listn=[2*n-1,2n];

如何将list1 * list2 * list3 * ... listn的结果作为笛卡尔积？

例如：

list1*list2*list3 should =[1,3,5],[1,3,6],[1,4,5],[1,4,6],[2,3,5],[2,3,6],[2,4,5],[2,4,6]

Answer 1

您可以使用递归来实现它，递归的基本情况是当输入为空然后返回空列表，否则处理剩余的元素。 E.g。

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

public class CartesianProduct {

    public static <T> List<List<T>> calculate(List<List<T>> input) {
        List<List<T>> res = new ArrayList<>();
        if (input.isEmpty()) { // if no more elements to process
            res.add(new ArrayList<>()); // then add empty list and return
            return res;
        } else {
            process(input, res); // we need to calculate the cartesian product of input and store it in res variable
        }
        return res; // method completes , return result
    }

    private static <T> void process(List<List<T>> lists, List<List<T>> res) {
        List<T> head = lists.get(0); //take first element of the list
        List<List<T>> tail = calculate(lists.subList(1, lists.size())); //invoke calculate on remaining element, here is recursion

        for (T h : head) { // for each head
            for (List<T> t : tail) { //iterate over the tail
                List<T> tmp = new ArrayList<>(t.size());
                tmp.add(h); // add the head
                tmp.addAll(t); // and current tail element
                res.add(tmp);
            }
        }
    }

    public static void main(String[] args) {
        //we invoke the calculate method
        System.out.println(calculate(Arrays.asList(Arrays.asList(1, 2), Arrays.asList(3, 4), Arrays.asList(5, 6))));
    }
}

输出

[[1, 3, 5], [1, 3, 6], [1, 4, 5], [1, 4, 6], [2, 3, 5], [2, 3, 6], [2, 4, 5], [2, 4, 6]]

Answer 2

感谢@ sol4me使用尾递归的答案，这是另一个不使用尾递归的版本，但我认为更容易理解。

public class CartesianProduct {

    public static <T> List<List<T>> calculate(List<List<T>> input) {
        List<List<T>> result= new ArrayList<List<T>>();

        if (input.isEmpty()) {  // If input a empty list
            result.add(new ArrayList<T>());// then add empty list and return
            return result;
        } else {
            List<T> head = input.get(0);//get the first list as a head
            List<List<T>> tail= calculate(input.subList(1, input.size()));//recursion to calculate a tail list
            for (T h : head) {//we merge every head element with every tail list.
                for (List<T> t : tail) {
                    List<T> resultElement = new ArrayList<T>();
                    resultElement.add(h);
                    resultElement.addAll(t);
                    result.add(resultElement);
                }
            }
        }
        return result;
    }


    public static void main(String[] args) {
        List<List<Integer>> bigList=Arrays.asList(Arrays.asList(1,2),Arrays.asList(3,4),Arrays.asList(5,6),Arrays.asList(7,8));
        System.out.println(calculate(bigList));
    }
}

Answer 3

使用嵌套循环的map-and-reduce方法

准备一个列表列表 List<List<T>> 填充了一个空值。该列表进一步用作中间结果的存储和最终结果。

将传入列表List<List<T>>中的数据依次追加到中间结果中，得到最终结果。示意性地，此迭代过程如下所示：

result0: [[]]
  list1: [1,2]
-------
result1: [[1],[2]]
  list2: [3,4]
-------
result2: [[1,3],[1,4],[2,3],[2,4]]
  list3: [5,6]
-------
result3: [[1,3,5],[1,3,6],[1,4,5],[1,4,6],[2,3,5],[2,3,6],[2,4,5],[2,4,6]]

Try it online!

/**
 * @param lists an arbitrary number of lists
 * @param <T>   the type of the elements
 * @return the Cartesian product
 */
public static <T> List<List<T>> cartesianProduct(List<List<T>> lists) {
    // check if incoming data is not null
    if (lists == null) return Collections.emptyList();
    // Cartesian product, intermediate result
    List<List<T>> cp = Collections.singletonList(Collections.emptyList());
    // iterate through incoming lists
    for (List<T> list : lists) {
        // non-null and non-empty lists
        if (list == null || list.size() == 0) continue;
        // intermediate result for next iteration
        List<List<T>> next = new ArrayList<>();
        // rows of current intermediate result
        for (List<T> row : cp) {
            // elements of current list
            for (T el : list) {
                // new row for next intermediate result
                List<T> nRow = new ArrayList<>(row);
                nRow.add(el);
                next.add(nRow);
            }
        }
        // pass to next iteration
        cp = next;
    }
    // Cartesian product, final result
    return cp;
}

public static void main(String[] args) {
    List<List<Integer>> lists = prepareLists(3);
    List<List<Integer>> cp = cartesianProduct(lists);
    // output without spaces
    System.out.println(lists.toString().replace(" ", ""));
    System.out.println(cp.toString().replace(" ", ""));
}

// supplementary method, prepares lists for multiplication
public static List<List<Integer>> prepareLists(int n) {
    List<List<Integer>> lists = new ArrayList<>(n);
    for (int i = 1; i <= n; i++)
        lists.add(Arrays.asList(i * 2 - 1, i * 2));
    return lists;
}

输出：

[[1,2],[3,4],[5,6]]
[[1,3,5],[1,3,6],[1,4,5],[1,4,6],[2,3,5],[2,3,6],[2,4,5],[2,4,6]]

^{另见：Generate all combinations from multiple lists}

如何从列表中获取笛卡尔积

3 个答案:

使用嵌套循环的map-and-reduce方法