78. Subsets

这里给出两种方法

方法一，利用上一题那个数字序列问题，把数字序列作为给定数组的下标来组成新数组，当元素填满时，将其中元素作为下标，换成nums中的对应元素即可

class Solution 
{
public:
    vector<vector<int>> subsets(vector<int>& nums) 
    {
        vector<vector<int>> res;
        vector<int> em;
        res.push_back(em);
        int sz=nums.size();
        for(int k=1;k<=sz;k++)
        {
            int i=0;
            vector<int>cur(k,0);
            while(i>=0)
            {
                cur[i]++;
                if(cur[i]>sz)
                    i--;
                else if(i==k-1)
                {
                    vector<int> curcopy=cur;
                    for(int &j:curcopy)
                        j=nums[j-1];
                    res.push_back(curcopy);
                }
                else
                {
                    i++;
                    cur[i]=cur[i-1];
                }
            }
        }
        return res;
    }
};

方法二，深度优先搜索，一级一级往下找，找到数组末尾时，返回即可

class Solution {
public:
    vector<vector<int>> subsets(vector<int>& nums) {
        vector<vector<int>> res;
        dfs(nums, 0, {}, res);
        return res;
    }
    void dfs(vector<int> &nums, int i, vector<int> cur, vector<vector<int>> &res) {
        if (i == nums.size()) {
            res.push_back(cur);
            return;   
        }
        dfs(nums, i+1, cur, res);
        cur.push_back(nums[i]);
        dfs(nums, i+1, cur, res);
    }
};

 方法三、回溯法

class Solution {
public:
    void backtrack(vector<vector<int>>& sets, vector<int>& temp, vector<int>& nums, int s)
    {
        sets.push_back(temp);
        for(int i = s; i < nums.size(); i++)
        {
            temp.push_back(nums[i]);
            backtrack(sets, temp, nums, i+1);
            temp.pop_back();
        }
    }
    
    vector<vector<int>> subsets(vector<int>& nums) {
        vector<vector<int>> sets;
        vector<int> temp;
        backtrack(sets, temp, nums, 0);
        return sets;
        
    }
};