90. Subsets II (Back-Track, DP)

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],
  []
]

思路: 对于重复了n次的字符，可以选择放入0,1,2...n个

class Solution {
public:
    vector<vector<int> > subsetsWithDup(vector<int> &S) {
        vector<vector<int>> result;
        vector<int> pre;
        if(S.size()==)
            return result;
        sort(S.begin(),S.end());
        result.push_back(pre);
        dfs(S,result,pre,);
        return result;
    }
    void dfs(vector<int> &S , vector<vector<int>> &result ,vector<int> pre , int depth)
    {
        if(depth == S.size())  return; //teminate condition

        int dupCounter = ;
        int dupNum = ;
        while(depth+ < S.size() && S[depth] == S[depth+]) //get duplicate times
        {
            depth++;
            dupNum++;
        }
        while(dupCounter++ <= dupNum) //push duplicate elements
        {
            pre.push_back(S[depth]);
            result.push_back(pre);
            dfs(S,result,pre,depth+);
        }
        dupCounter = ;
        while(dupCounter++ <= dupNum) //backtracking
        {
            pre.pop_back();
        }
        dfs(S, result,pre, depth+); //push none, dfs directly
    }
};

思路II：DP，插入排序法增加元素。重复的元素要在一个for循环内插入，否则会导致subset有重复。

class Solution {
public:
    vector<vector<int>> subsetsWithDup(vector<int>& nums) {
        vector<vector<int>> ret;
        vector<int> retItem;
        ret.push_back(retItem);
        int size; //number of memebers in ret
        int count = ; //count the duplicate number

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

        for(int i = ; i < nums.size(); i++){ //iterate the number to insert
            if(i < nums.size()- && nums[i+]==nums[i]){
                count++;
                continue;
            }

            size = ret.size();
            for(int j = ; j < size; j++){ //iterate current item in ret
                vector<int> newItem = ret[j];
                for(int k = ; k < count; k++){ //duplicate 1,2,...,count times
                    newItem.push_back(nums[i]);
                    ret.push_back(newItem);
                }
            }
            count = ;
        }
        return ret;
    }
};