[LeetCode] Subsets II 子集

Given a collection of integers that might contain duplicates, S, return all possible subsets.
Note:

1. Elements in a subset must be in non-descending order.
2. The solution set must not contain duplicate subsets.

For example, if S = [1,2,2], a solution is:

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

如果集合S里没有重复的数字，那么每个数字可能出现在subset，也可能不出现；如果集合S里有重复的数字，假设这个重复数字的个数是4，那么这个重复数字在subset里出现的次数可能是0、1、2、3、4. 根据这个观察，就可以写出如下的代码。

    vector<vector<int> > subsetsWithDup(vector<int> &S) {
        
        sort(S.begin(), S.end()); // sort the input vector into increasing order
        
        vector<int> nums; // the unique set of numbers in S
        vector<int> counts; // the count of each number in nums
        // If input S=[1,2,2,3,3,3,3], then nums=[1,2,3], counts=[1,2,4]
        
        for(int i=0; i<S.size(); i++)
        {
            if(i==0)
            {
                nums.push_back(S[0]);
                counts.push_back(1);
            }
            else
            {
                if(S[i]==S[i-1])
                    counts[counts.size()-1]++;
                else
                {
                    nums.push_back(S[i]);
                    counts.push_back(1);
                }
            }
        }
        
        vector<int> result;
        vector<vector<int> > results;
        subsetsWithDupUtil(nums, counts, result, 0, results);
        
        return results;
        
    }
    
    void subsetsWithDupUtil(vector<int> & nums, vector<int>& counts, vector<int> result, int pos, vector<vector<int> >& results)
    {
        if(pos==nums.size())
        {
            results.push_back(result);
            return;
        }
        
        for(int i=0; i<=counts[pos]; i++)
        {
            subsetsWithDupUtil(nums, counts, result, pos+1, results);
            result.push_back(nums[pos]);
        }
    }