Given a list of numbers that may has duplicate numbers, return all possible subsets

Notice

• Each element in a subset must be in non-descending order.
• The ordering between two subsets is free.
• The solution set must not contain duplicate subsets.

If S = [1,2,2], a solution is:

[
  [2],
  [1],
  [1,2,2],
  [2,2],
  [1,2],
  []
]

Can you do it in both recursively and iteratively?

題意

給定一個可能具有重復數字的列表，返回其所有可能的子集

註意事項

• 子集中的每個元素都是非降序的
• 兩個子集間的順序是無關緊要的
• 解集中不能包含重復子集

解法一：

 1 class Solution {
 2 public:
 3     /*
 4      * @param nums: A set of numbers.
 
 5      * @return: A list of lists. All valid subsets.
 6      */
 7     vector<vector<int>> subsetsWithDup(vector<int> &nums) {
 8         // write your code here
 9         vector<vector<int> > results;
10         vector<int> result;
11         
12         sort(nums.begin(), nums.end());
 
13         
14         helper(nums, 0, result, results);
15         
16         return results;
17     }
18     
19     void helper(vector<int> &nums, int start, vector<int> & result, vector<vector<int> > & results)
20     {
21         results.push_back(result);
22         
23         for (int i = start; i < nums.size(); ++i) {
24             if (i > start && nums[i] == nums[i - 1]) {
25                 continue;
26             }
27             
28             result.push_back(nums[i]);
29             
30             helper(nums, i + 1, result, results);
31             
32             result.pop_back();
33         }
34     }
35 };

18. Subsets II【medium】