[LeetCode]Permutations II

题目描述：

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].

解题思路：

思路和上题目相似，但需要跳过重复的元素

class Solution {
public:
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        vector<vector<int>> result;
        vector<int> elem;
        vector<bool> visited(nums.size(), false);
        permute(nums, visited, result, elem);
        return result;
    }
private:
    void permute(const vector<int> &nums, vector<bool> &visited, vector<vector<int>> &result, vector<int> &elem) {
        if (nums.size() == elem.size()) {
            result.push_back(elem);
            return;
        }
        
        for (int i = 0; i < nums.size(); ++i) {
            if (!visited[i]) {
                visited[i] = true;
                elem.push_back(nums[i]);
                permute(nums, visited, result, elem);
                elem.pop_back();
                visited[i] = false;
                while (i < nums.size() - 1 && nums[i] == nums[i + 1]) ++i; // 跳过重复的元素
            }
        }
    }
};