047. Permutations II

Difficulty: Medium

Frequency: N/A

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].


My solution:
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Code:
class Solution {
public:
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        vector<vector<int>> results;
        vector<int> tmp;
        vector<bool> used(nums.size(), false);
        sort(nums.begin(), nums.end());
        permuteHelper(nums, used, tmp, results);
        return results;
    }
    void permuteHelper(vector<int> nums, vector<bool> used, vector<int> tmp, vector<vector<int>> &results) {
        if(tmp.size() == nums.size()) {
            results.push_back(tmp);
            return;
        }
        for (int i = 0; i < nums.size(); i++) {
            if(used[i]) continue;
            if (i > 0 && nums[i] == nums[i - 1] && !used[i - 1]) continue;
            used[i] = true;
            tmp.push_back(nums[i]);
            permuteHelper(nums, used, tmp, results);
            tmp.pop_back();
            used[i] = false;
        }
    }
};

Another solution:
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steps:
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Things to learn:
1. How to skip duplicate element
2. Avoid using lookup from previous result set.
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089. Gray Code

Difficulty: Medium

Frequency: N/A

The gray code is a binary numeral system where two successive values differ in only one bit.

Given a non-negative integer n representing the total number of bits in the code, print the sequence of gray code. A gray code sequence must begin with 0.

For example, given n = 2, return [0,1,3,2]. Its gray code sequence is:

00 - 0
01 - 1
11 - 3
10 - 2

Note:
For a given n, a gray code sequence is not uniquely defined.

For example, [0,2,3,1] is also a valid gray code sequence according to the above definition.

For now, the judge is able to judge based on one instance of gray code sequence. Sorry about that.


My solution:
Data structure:
vector
Steps:
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Code:
class Solution {
public:
    vector<int> grayCode(int n) {
        vector<int> result;
        result.push_back(0);
        for (int i = 0; i < n; i++) {
            int highBit = 1 << i;
            for (int j = result.size() - 1; j >= 0; j--) {
                result.push_back(highBit + result[j]);
            }
        }
        return result;
    }
};

Another solution:
Data structure:
steps:
Complexity:
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Things to learn:
1. How to do it in backtracking way?