47. Permutations II

description:

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

You can assume that you can always reach the last index.

Example:

Example:

Input: [1,1,2]
Output:
[
  [1,1,2],
  [1,2,1],
  [2,1,1]
]

answer:

solution1:

class Solution {
public:
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        if (nums.empty()) return {{}};
        set<vector<int>> res;
        int first = nums[0];
        nums.erase(nums.begin());
        vector<vector<int>> words = permuteUnique(nums);
        for(auto &a : words) {
            for (int i = 0; i <= a.size(); i++) {
                a.insert(a.begin() + i, first);
                res.insert(a); // 这里 set 只能用 insert，不能和 vector 一样用 push_back
                a.erase(a.begin() + i);
            }
        }
        return vector<vector<int>> (res.begin(), res.end()); // 这里不能直接返回 res, 因为题目要求是 vector<vector<int>>, not set<vector>
    }
};

solution2:

class Solution {
public:
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        vector<vector<int>> res;
        vector<int> out, visited(nums.size(), 0);
        sort(nums.begin(), nums.end());
        permuteDFS(nums, 0, visited, out, res);
        return res;
    }
    void permuteDFS(vector<int>& nums, int level, vector<int>& visited, vector<int>& out, vector<vector<int>>& res) {
        if (level >= nums.size()) {res.push_back(out); return;}
        for (int i = 0; i < nums.size(); ++i) {
            if (visited[i] == 1) continue;
            if (i > 0 && nums[i] == nums[i - 1] && visited[i - 1] == 0) continue; // key
            visited[i] = 1;
            out.push_back(nums[i]);
            permuteDFS(nums, level + 1, visited, out, res);
            out.pop_back();
            visited[i] = 0;
        }
    }
};  

relative point get√：

hint :

第二种方法，在第一个1结束递归后，会置回0，这个时候第二个1就会因为那个if语句而跳过去，那么就避免了和之前那个1一模一样的一次递归。也就是说，一样的数字，在最开始的放置在第一位不要重复那么后面就不会重复了。