【Leetcode】Permutations II

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

For example,
[1,1,2] have the following unique permutations:
[1,1,2], [1,2,1], and [2,1,1].

class Solution {
public:
    vector<vector<int> > permuteUnique(vector<int> &num) {
        vector<vector<int> > result;
        sort(num.begin(), num.end());
        vector<bool> used(100, false);
        vector<int> path;
        dfs(num, path, used, result);
        return result;
    }
    
    private:
    void dfs(vector<int> &num, vector<int> &path, vector<bool> used, vector<vector<int>> &result) {
        if (path.size() == num.size()) {
            result.push_back(path);
            return;
        }
        for (size_t i = 0; i < num.size(); ++i) {
            if (used[i] == false && (i == 0 || num[i] > num[i - 1] || used[i - 1])) {
                path.push_back(num[i]);
                used[i] = true;
                dfs(num, path, used, result);
                used[i] = false;
                path.pop_back();
            }
        } 
    }
};

与上一题类似，因为数字可能不同需要的处理有：用一个表记录下各位数有没有被用过，当遇到相同数字的时候，只有排在它之前的相等的数字都被用过了才选它(相当于赋予了编号)。