Refer to上一篇
https://blog.csdn.net/weixin_43476349/article/details/83989562
这类题的共同点都是用backtracking，建立一个helper API然后recursion

• input nums needs sort，为了方便remove duplicates
  sort(nums.begin(), nums.end())

helper API有几个共同点：
（跟combinations不一样的是）
需要一个tovisit来标记被访问过的元素
loop每次都是从i = 0开始的，但是combinations loop是从start开始的。
所以：

• input：res，row，tovisit （没有start）
• exit condition：
  return都是array size要求，exit condition就是：row.size（）== nums.size()
• 如果input candidates有重复元素，需要remove duplicates，
  while(i+1 < candidates.size() && candidates[i] == candidates[i+1]) i++;
  比如permutation II。

46. Permutations
    Given a collection of distinct integers, return all possible permutations.

class Solution {
public:
    vector<vector<int>> permute(vector<int>& nums) {
        vector<vector<int>>res;
        vector<int>row;
        vector<int>tovisit(nums.size(), 0);
        helper(res, row, tovisit, nums);
        return res;
        
    }
    
    void helper(vector<vector<int>>&res, vector<int>&row, vector<int>&tovisit, vector<int>& nums)
    {
        if(row.size() == nums.size())
        {
            res.push_back(row);
            return;
        }
        
        for(int i = 0; i < nums.size(); i++)
        {
            if(tovisit[i] == 0)
            {
                tovisit[i] = 1;
                row.push_back(nums[i]);
                helper(res, row, tovisit, nums);
                row.pop_back();
                tovisit[i] = 0;
            }
        }
    }
};

47. Permutations II
    Given a collection of numbers that might contain duplicates, return all possible unique permutations.

class Solution {
public:
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        vector<vector<int>>res;
        vector<int>row;
        vector<int>tovisit(nums.size(), 0);
        sort(nums.begin(), nums.end());
        helper(res, row, tovisit, nums);
        return res;
        
    }
    
    void helper(vector<vector<int>>&res, vector<int>&row, vector<int>tovisit, vector<int>& nums)
    {
        if(row.size() == nums.size())
        {
            res.push_back(row);
            return;
        }
        
        for(int i = 0; i < nums.size(); i++)
        {
            if(tovisit[i] == 0)
            {
                tovisit[i] = 1;
                row.push_back(nums[i]);
                helper(res, row, tovisit, nums);
                row.pop_back();
                tovisit[i] = 0;     
                while(i + 1 < nums.size() && nums[i] == nums[i+1]) i++;
            }
            
        }
    }
};