18. Subsets II【medium】

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.

Example

If S = [1,2,2], a solution is:

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

Challenge

Can you do it in both recursively and iteratively?

题意

给定一个可能具有重复数字的列表，返回其所有可能的子集

注意事项

• 子集中的每个元素都是非降序的
• 两个子集间的顺序是无关紧要的
• 解集中不能包含重复子集

解法一：

 class Solution {
 public:
     /*
      * @param nums: A set of numbers.
      * @return: A list of lists. All valid subsets.
      */
     vector<vector<int>> subsetsWithDup(vector<int> &nums) {
         // write your code here
         vector<vector<int> > results;
         vector<int> result;

         sort(nums.begin(), nums.end());

         helper(nums, , result, results);

         return results;
     }

     void helper(vector<int> &nums, int start, vector<int> & result, vector<vector<int> > & results)
     {
         results.push_back(result);

         for (int i = start; i < nums.size(); ++i) {
             if (i > start && nums[i] == nums[i - ]) {
                 continue;
             }

             result.push_back(nums[i]);

             helper(nums, i + , result, results);

             result.pop_back();
         }
     }
 };