给定一个数组，如何生成子集大小k的所有组合? [英] Given an array, how to generate all combinations of subset size k?

问题描述

因此，给定input = [1, 2, 3]和k=2，这将返回:

So given input = [1, 2, 3] and k=2 this would return:

1 2
1 3
2 1
2 3
3 1
3 2

这是我所寻找的最接近的内容，但并非完全相同: http://algorithms.tutorialhorizo​​n.com/print-all-combinations-of-subset-of-size-k-from-given-array/

This is the closest to what I am looking for, but not quite: http://algorithms.tutorialhorizon.com/print-all-combinations-of-subset-of-size-k-from-given-array/

function subsetsOfSize(a, used, startIndex, currentSize, k) {
  if (currentSize === k) {
    for (var i = 0; i < a.length; i++) {
      if (used[i])
        console.log(a[i]);
    }
    console.log('-');
    return;
  }
    
  if (startIndex === a.length)
    return;
    
  used[startIndex] = true;
  subsetsOfSize(a, used, startIndex+1, currentSize+1, k);

  used[startIndex] = false;
  subsetsOfSize(a, used, startIndex+1, currentSize, k);
}

var input = [1,2,3];
subsetsOfSize(input, Array(input.length).fill(false), 0, 0, 2);

^缺少诸如2 1，3 1，3 2等的结果.

^ Missing results such as 2 1, 3 1, 3 2, etc.

其次，我不确定我是否正确地命名了此问题，因为大小为k的子集的所有组合"的解决方案没有给出预期的答案.

Secondly, I am not sure if I am naming this problem correctly because solutions to "all combinations of subset of size k" do not give the expected answer.

推荐答案

一种递归解决方案，用于查找k个子集的排列(以伪代码表示):

A recursive solution to find k-subset permutations (in pseudo-code):

kSubsetPermutations(partial, set, k) {
    for (each element in set) {
        if (k equals 1) {
            store partial + element
        }
        else {
            make copy of set
            remove element from copy of set
            recurse with (partial + element, copy of set, k - 1)
        }
    }
}

下面是一个示例:

输入:[a，b，c，d，e]
k:3

input: [a,b,c,d,e]
k: 3

partial = [], set = [a,b,c,d,e], k = 3
    partial = [a], set = [b,c,d,e], k = 2
        partial = [a,b], set = [c,d,e], k = 1 -> [a,b,c], [a,b,d], [a,b,e]
        partial = [a,c], set = [b,d,e], k = 1 -> [a,c,b], [a,c,d], [a,c,e]
        partial = [a,d], set = [b,c,e], k = 1 -> [a,d,b], [a,d,c], [a,d,e]
        partial = [a,e], set = [b,c,d], k = 1 -> [a,e,b], [a,e,c], [a,e,d]
    partial = [b], set = [a,c,d,e], k = 2
        partial = [b,a], set = [c,d,e], k = 1 -> [b,a,c], [b,a,d], [b,a,e]
        partial = [b,c], set = [a,d,e], k = 1 -> [b,c,a], [b,c,d], [b,c,e]
        partial = [b,d], set = [a,c,e], k = 1 -> [b,d,a], [b,d,c], [b,d,e]
        partial = [b,e], set = [a,c,d], k = 1 -> [b,e,a], [b,e,c], [b,e,d]
    partial = [c], set = [a,b,d,e], k = 2
        partial = [c,a], set = [b,d,e], k = 1 -> [c,a,b], [c,a,d], [c,a,e]
        partial = [c,b], set = [a,d,e], k = 1 -> [c,b,a], [c,b,d], [c,b,e]
        partial = [c,d], set = [a,b,e], k = 1 -> [c,d,a], [c,d,b], [c,d,e]
        partial = [c,e], set = [a,b,d], k = 1 -> [c,e,a], [c,e,b], [c,e,d]
    partial = [d], set = [a,b,c,e], k = 2
        partial = [d,a], set = [b,c,e], k = 1 -> [d,a,b], [d,a,c], [d,a,e]
        partial = [d,b], set = [a,c,e], k = 1 -> [d,b,a], [d,b,c], [d,b,e]
        partial = [d,c], set = [a,b,e], k = 1 -> [d,c,a], [d,c,b], [d,c,e]
        partial = [d,e], set = [a,b,c], k = 1 -> [d,e,a], [d,e,b], [d,e,c]
    partial = [e], set = [a,b,c,d], k = 2
        partial = [e,a], set = [b,c,d], k = 1 -> [e,a,b], [e,a,c], [e,a,d]
        partial = [e,b], set = [a,c,d], k = 1 -> [e,b,a], [e,b,c], [e,b,d]
        partial = [e,c], set = [a,b,d], k = 1 -> [e,c,a], [e,c,b], [e,c,d]
        partial = [e,d], set = [a,b,c], k = 1 -> [e,d,a], [e,d,b], [e,d,c]

function kSubsetPermutations(set, k, partial) {
    if (!partial) partial = [];                 // set default value on first call
    for (var element in set) {
        if (k > 1) {
            var set_copy = set.slice();         // slice() creates copy of array
            set_copy.splice(element, 1);        // splice() removes element from array
            kSubsetPermutations(set_copy, k - 1, partial.concat([set[element]]));
        }                                       // a.concat(b) appends b to copy of a
        else document.write("[" + partial.concat([set[element]]) + "] ");
    }
}
kSubsetPermutations([1,2,3,4,5], 3);

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