因此，考虑input = [1, 2, 3]和k=2将返回：
input = [1, 2, 3]
k=2
1 2 1 3 2 1 2 3 3 1 3 2
这是最接近我要寻找的内容，但不太准确：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);
^缺少的结果，如2 1，3 1，3 2等。
2 1
3 1
3 2
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。
查找k个子集置换的递归解决方案（用伪代码）：
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) } } }
这是一个示例示例：
输入：[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);
