我找到了similar question on SO 并在 MATLAB 中实现了建议的解决方案:
function [result] = kth_combination(k,l,r)
if r == 0
result = [];
elseif size(l,2) == r
result = l;
else
i = nchoosek(size(l,2)-1,r-1);
if k < i+1
result = [l(1), kth_combination(k, l(2:end), r-1)];
else
result = kth_combination(k-i, l(2:end), r);
end
end
end
在 MATLAB File Exchange 上还有另一种解决方案,它不是基于递归:onecomb
为了比较这 3 个解决方案,我创建了这个基准函数:
function [time_1,time_2, time_3] = compare_solutions(m,n,i,num_runs)
time_1 = 0;
time_2 = 0;
time_3 = 0;
for run=1:num_runs
tic
A=nchoosek(1:m,n);
res_1 = A(i,:);
time_1 = time_1 + toc;
tic
res_2 = kth_combination(i,1:m,n);
time_2 = time_2 + toc;
tic
res_3 = onecomb(m,n,i);
time_3 = time_3 + toc;
if (run==1) && (sum(res_1 ~= res_2) || sum(res_1 ~= res_3))
error('solutions are NOT identical');
end
end
time_1 = time_1/num_runs;
time_2 = time_2/num_runs;
time_3 = time_3/num_runs;
end
示例运行:
>> [time_1,time_2, time_3] = compare_solutions(20,10,10,10)
time_1 =
1.9676
time_2 =
6.8508e-04
time_3 =
7.1848e-05
第二种和第三种解决方案比 nchoosek 方法快得多,非递归方法比递归方法快 10 倍。