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, 0, 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 - 1]) {
                continue;
            }
            
            result.push_back(nums[i]);
            
            helper(nums, i + 1, result, results);
            
            result.pop_back();
        }
    }
};