从数组 (Java) 中获取大小为 n 的所有组合的算法？

Question

现在我正在尝试编写一个函数，它接受一个数组和一个整数 n，并给出每个大小 n 组合的列表（因此是一个 int 数组列表）。我可以使用 n 个嵌套循环来编写它，但这只适用于特定大小的子集。我无法弄清楚如何将它概括为适用于任何规模的组合。我想我需要使用递归？

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。

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");
        }
    }
}

Answer 1

你绝对可以通过迭代来做到这一点。

这是一个解决方案，它计算我们应该创建多少个数组，然后使用数学构建它们以计算源数组中的哪一项应该放在什么地方：

public static void combinations(int n, int[] arr, List<int[]> list) {
    // Calculate the number of arrays we should create
    int numArrays = (int)Math.pow(arr.length, n);
    // Create each array
    for(int i = 0; i < numArrays; i++) {
        int[] current = new int[n];
        // Calculate the correct item for each position in the array
        for(int j = 0; j < n; j++) {
            // This is the period with which this position changes, i.e.
            // a period of 5 means the value changes every 5th array
            int period = (int) Math.pow(arr.length, n - j - 1);
            // Get the correct item and set it
            int index = i / period % arr.length;
            current[j] = arr[index];
        }
        list.add(current);
    }
}

更新：

这是一个优化版本，可以显着减少对 Math.pow

的调用次数

public static void combinations(int n, int[] arr, List<int[]> list) {
    // Calculate the number of arrays we should create
    int numArrays = (int)Math.pow(arr.length, n);
    // Create each array
    for(int i = 0; i < numArrays; i++) {
        list.add(new int[n]);
    }
    // Fill up the arrays
    for(int j = 0; j < n; j++) {
        // This is the period with which this position changes, i.e.
        // a period of 5 means the value changes every 5th array
        int period = (int) Math.pow(arr.length, n - j - 1);
        for(int i = 0; i < numArrays; i++) {
            int[] current = list.get(i);
            // Get the correct item and set it
            int index = i / period % arr.length;
            current[j] = arr[index];
        }
    }
}

Answer 2

如果我对你的问题的理解正确，this 文章似乎指出了你正在尝试做的事情。

引用文章：

Method 1 (Fix Elements and Recur)

We create a temporary array ‘data[]’ which stores all outputs one by one. The idea is to start from first index (index = 0) in data[], one by one fix elements at this index and recur for remaining indexes. Let the input array be {1, 2, 3, 4, 5} and r be 3. We first fix 1 at index 0 in data[], then recur for remaining indexes, then we fix 2 at index 0 and recur. Finally, we fix 3 and recur for remaining indexes. When number of elements in data[] becomes equal to r (size of a combination), we print data[].

Method 2 (Include and Exclude every element)

Like the above method, We create a temporary array data[]. The idea here is similar to Subset Sum Problem. We one by one consider every element of input array, and recur for two cases:

The element is included in current combination (We put the element in data[] and increment next available index in data[])

The element is excluded in current combination (We do not put the element and do not change index)

When number of elements in data[] become equal to r (size of a combination), we print it.

Answer 3

这是一个经过充分研究的生成所有 k 子集的问题，或 k-combinations，无需递归即可轻松完成。

这个想法是让大小为 k 的数组保持输入数组中元素的索引的序列（它们是从 0 到 n - 1) 按升序排列。（子集然后可以通过从初始数组中获取这些索引的项目来创建。）所以我们需要生成所有这样的索引序列。

第一个索引序列将是[0, 1, 2, ... , k - 1]，在第二步切换到[0, 1, 2,..., k]，然后是[0, 1, 2, ... k + 1]等等。最后可能的序列将是 [n - k, n - k + 1, ..., n - 1].

在每一步中，算法都会寻找最接近可以递增的最终项目，将其递增并填充该项目的项目。

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......在某个时候我们有 [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] 之后是 [1, 2, 3]，然后是 [1, 2, 4] 等等

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;
}

从数组 (Java) 中获取大小为 n 的所有组合的算法？

Algorithm to get all the combinations of size n from an array (Java)?

java

combinations

combinatorics