Given a collection of integers that might contain duplicates, nums, return all possible subsets.
Note:
For example,
If nums = [1,2,2], a solution is:
[ [2], [1], [1,2,2], [2,2], [1,2], [] ]
区别在于这边需要临时搭一个数组的复制，单次只能去掉一个元素而不是使用 ans - [x] 去除ans内所有为x的元素。
def subsets_with_dup(nums) sub([nums.sort],0) end def sub(ans,start=0) t = ans.length (start...t).each do |i| ans[i].each do |x| temp = ans[i].dup temp.length.times do |j| if temp[j] == x temp.delete_at(j) break end end ans << temp unless ans.include?(temp) end end return ans if ans[-1].empty? sub(ans,t) end
原文：http://www.cnblogs.com/lilixu/p/4640598.html