题目

Given a collection of integers that might contain duplicates, nums, return all possible subsets.

Note:
Elements in a subset must be in non-descending order.
The solution set must not contain duplicate subsets.
For example,
If nums = [1,2,2], a solution is:

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

分析

求带有重复元素的序列的全子集；

用动态规划的思想，逐个向前i-1的元素的子集中添加第i个元素，添加时需要判重，若已存在，则不添加。

AC代码

class Solution {
public:
    vector<vector<int>> subsetsWithDup(vector<int>& nums) {
        vector<vector<int> > ret(1, vector<int>());

        if (nums.empty())
            return ret;

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

        int size = nums.size();
        for (int i = 0; i < size; ++i)
        {
            ret = subSets(ret, nums, i);
        }
        return ret;
    }

    vector<vector<int> > subSets(vector<vector<int> > &ret, vector<int> &nums, int &idx)
    {
        vector<int> tmp;
        int count = ret.size();

        //对于每一个已有子集合，加入新元素
        for (int i = 0; i < count; ++i)
        {
            //当前集合
            tmp = ret[i];       
            tmp.push_back(nums[idx]);
            if (find(ret.begin(), ret.end(), tmp) != ret.end())
                continue;
            ret.push_back(tmp); 
        }//for

        return ret;
    }
};

GitHub测试程序源码