从数组中获取大小为n的所有组合的算法（Java）？ [英] Algorithm to get all the combinations of size n from an array (Java)?

问题描述

现在我正在尝试编写一个接受数组和整数n的函数，并给出每个大小为n的组合的列表（所以是int数组的列表）。我能够使用n个嵌套循环编写它，但这仅适用于特定大小的子集。我无法弄清楚如何推广它适用于任何大小的组合。我想我需要使用递归？

Right now I'm trying to write a function that takes an array and an integer n, and gives a list of each size n combination (so a list of int arrays). I am able to write it using n nested loops, but this only works for a specific size of subset. I can't figure out how to generalize it to work for any size of combination. I think I need to use recursion?

这是3个元素的所有组合的代码，我需要一个适用于任意数量元素的算法。

This is the code for all combinations of 3 elements, and I need an algorithm for any number of elements.

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

public class combinatorics{
    public static void main(String[] args) {

        List<int[]> list = new ArrayList<int[]>();
        int[] arr = {1,2,3,4,5};
        combinations3(arr,list);
        listToString(list);
    }

    static void combinations3(int[] arr, List<int[]> list){
        for(int i = 0; i<arr.length-2; i++)
            for(int j = i+1; j<arr.length-1; j++)
                for(int k = j+1; k<arr.length; k++)
                    list.add(new int[]{arr[i],arr[j],arr[k]});
    }

    private static void listToString(List<int[]> list){
        for(int i = 0; i<list.size(); i++){ //iterate through list
            for(int j : list.get(i)){ //iterate through array
                System.out.printf("%d ",j);
            }
        System.out.print("\n");
        }
    }
}

推荐答案

这是一个生成所有k子集的充分研究的问题，或 k-combinations ，这可以在没有递归的情况下轻松完成。

This is a well-studied problem of generating all k-subsets, or k-combinations, which can be easily done without recursion.

我们的想法是让数组大小 k 保持序列输入数组中元素的 indices （从 0 到 n - 1 的数字）按顺序递增。 （ Subset 然后可以通过从初始数组中通过这些索引获取项目来创建。）因此我们需要生成所有这样的索引序列。

The idea is to have array of size k keeping sequence of indices of elements from the input array (which are numbers from 0 to n - 1) in increasing order. (Subset then can be created by taking items by these indices from the initial array.) So we need to generate all such index sequences.

第一个索引序列将是 [0,1,2，...，k - 1] ，第二步它切换到 [0， 1,2，...，k] ，然后到 [0,1,2，... k + 1] 等等。最后一个可能的序列是 [n - k，n - k + 1，...，n - 1] 。

First index sequence will be [0, 1, 2, ... , k - 1], on the second step it switches to [0, 1, 2,..., k], then to [0, 1, 2, ... k + 1] and so on. The last possible sequence will be [n - k, n - k + 1, ..., n - 1].

在每一步中，算法会查找最接近最终项目的值，该项目可以递增，递增并将项目填充到该项目。

On each step, algorithm looks for the closest to the end item which can be incremented, increments it and fills up items right to that item.

为了说明，请考虑 n = 7 和 k = 3 。第一个索引序列是 [0,1,2] ，然后是 [0,1,3] 等等。 ..在某些时候我们有 [0,5,6] ：

To illustrate, consider n = 7 and k = 3. First index sequence is [0, 1, 2], then [0, 1, 3] and so on... At some point we have [0, 5, 6]:

[0, 5, 6] <-- scan from the end: "6" cannot be incremented, "5" also, but "0" can be
[1, ?, ?] <-- "0" -> "1"
[1, 2, 3] <-- fill up remaining elements

next iteration:

[1, 2, 3] <-- "3" can be incremented
[1, 2, 4] <-- "3" -> "4"

因此， [0,5,6] 后跟 [1,2,3] ，然后去 [1,2,4] 等等。

Thus, [0, 5, 6] is followed by [1, 2, 3], then goes [1, 2, 4] etc.

代码：

int[] input = {10, 20, 30, 40, 50};    // input array
int k = 3;                             // sequence length   

List<int[]> subsets = new ArrayList<>();

int[] s = new int[k];                  // here we'll keep indices 
                                       // pointing to elements in input array

if (k <= input.length) {
    // first index sequence: 0, 1, 2, ...
    for (int i = 0; (s[i] = i) < k - 1; i++);  
    subsets.add(getSubset(input, s));
    for(;;) {
        int i;
        // find position of item that can be incremented
        for (i = k - 1; i >= 0 && s[i] == input.length - k + i; i--); 
        if (i < 0) {
            break;
        }
        s[i]++;                    // increment this item
        for (++i; i < k; i++) {    // fill up remaining items
            s[i] = s[i - 1] + 1; 
        }
        subsets.add(getSubset(input, s));
    }
}

// generate actual subset by index sequence
int[] getSubset(int[] input, int[] subset) {
    int[] result = new int[subset.length]; 
    for (int i = 0; i < subset.length; i++) 
        result[i] = input[subset[i]];
    return result;
}

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