78. Subsets (Back-Track, DP)

Given a set of distinct integers, 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 = [1,2,3], a solution is:

[
  [3],
  [1],
  [2],
  [1,2,3],
  [1,3],
  [2,3],
  [1,2],
  []
]

思路：遍历数组，有选择当前元素和不选当前元素两种操作，所以使用带回溯的递归

class Solution {
public:
    vector<vector<int>> subsets(vector<int> &S) {
        vector<vector<int>> result;
        vector<int> pre;
        result.push_back(pre);
        if(S.size()==)
            return result;
        sort(S.begin(),S.end());
        DepthFirst(S,result,pre,);
        return result;
    }
    void DepthFirst(vector<int> &S , vector<vector<int>> &result ,vector<int> &pre , int depth)
    {
        if (depth==S.size()) {
            return;
        }
        pre.push_back(S[depth]);
        result.push_back(pre);
        DepthFirst(S,result,pre,depth+);
        pre.pop_back();
        DepthFirst(S,result,pre,depth+);
    }
};

思路II: DP. n个元素的数组结果，是n-1个数组结果，再加上插入第n个元素的结果

class Solution {
public:
    vector<vector<int>> subsets(vector<int>& nums) {
        vector<vector<int>> ret;
        vector<int> retItem;
        ret.push_back(retItem);
        int size; //number of memebers in ret
        for(int i = ; i < nums.size(); i++){ //iterate the number to insert
            size = ret.size();
            for(int j = ; j < size; j++){ //iterate current item in ret
                vector<int> newItem = ret[j];
                newItem.push_back(nums[i]);
                ret.push_back(newItem);
            }
        }
        return ret;
    }
};