## 题目地址 
https://leetcode-cn.com/problems/permutations-ii/

## 思路 

i > 0 && nums[i] == nums[i-1] && used[i-1] == false

这是最高效的，可以用 1 1 1 1 1  跑一个样例试试

## C++代码

```
class Solution {
private:
    vector<vector<int>> result;
    void backtracking (vector<int>& nums, vector<int>& vec, vector<bool>& used) {
        // 此时说明找到了一组
        if (vec.size() == nums.size()) {
            result.push_back(vec);
            return;
        }

        for (int i = 0; i < nums.size(); i++) {
            if (i > 0 && nums[i] == nums[i-1] && used[i-1] == false) {
                continue;
            }
            if (used[i] == false) {
                used[i] = true;
                vec.push_back(nums[i]);
                backtracking(nums, vec, used);
                vec.pop_back();
                used[i] = false;
            }
        }
    }

public:
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        vector<bool> used(nums.size(), false);
        vector<int> vec;
        backtracking(nums, vec, used);
        return result;

    }
};
```