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>& nums) {
vector<vector<int>> paths;
if (nums.empty()) {
return paths;
} vector<int> index;
vector<int> path;
permuteUniqueHelper(nums, index, path, paths);
return paths; } private:
void permuteUniqueHelper(const vector<int> &nums,
vector<int> &index,
vector<int> &path,
vector<vector<int>> &paths) {
if (path.size() == nums.size()) {
paths.push_back(path);
return;
} // 保证相同的数不在同一位置出现两次以上
unordered_set<int> hashset;
for (int ix = 0; ix < nums.size(); ix++) {
if (find(index.begin(), index.end(), ix) == index.end() && hashset.count(nums[ix]) == 0) {
hashset.insert(nums[ix]); index.push_back(ix);
path.push_back(nums[ix]);
permuteUniqueHelper(nums, index, path, paths);
index.pop_back();
path.pop_back();
}
}
}
};

能否对空间复杂度做进一步的优化?

class Solution {
public:
/**
* @param nums: A list of integers.
* @return: A list of unique permutations.
*/
vector<vector<int> > permuteUnique(vector<int> &nums) {
// write your code here
vector<vector<int>> paths;
if (nums.empty()) {
return paths;
} sort(nums.begin(), nums.end());
bool *visited = new bool[nums.size()]();
vector<int> path;
permuteUniqueHelper(nums, visited, path, paths);
return paths;
} private:
void permuteUniqueHelper(const vector<int> &nums,
bool visited[],
vector<int> &path,
vector<vector<int>> &paths) {
if (path.size() == nums.size()) {
paths.push_back(path);
return;
} for (int ix = 0; ix < nums.size(); ix++) {
if (visited[ix] == true || ix > 0 && nums[ix - 1] == nums[ix] && visited[ix - 1] == false) {
continue;
} visited[ix] = true;
path.push_back(nums[ix]);
permuteUniqueHelper(nums, visited, path, paths);
visited[ix] = false;
path.pop_back();
}
}
};

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