LeetCode Solutions
47. Permutations II
Time: $O(n \cdot n!)$ Space: $O(n \cdot n!)$
class Solution {
public:
vector<vector<int>> permuteUnique(vector<int>& nums) {
vector<vector<int>> ans;
sort(begin(nums), end(nums));
dfs(nums, vector<bool>(nums.size()), {}, ans);
return ans;
}
private:
void dfs(const vector<int>& nums, vector<bool>&& used, vector<int>&& path,
vector<vector<int>>& ans) {
if (path.size() == nums.size()) {
ans.push_back(path);
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;
path.push_back(nums[i]);
dfs(nums, move(used), move(path), ans);
path.pop_back();
used[i] = false;
}
}
};
class Solution {
public List<List<Integer>> permuteUnique(int[] nums) {
List<List<Integer>> ans = new ArrayList<>();
Arrays.sort(nums);
dfs(nums, new boolean[nums.length], new ArrayList<>(), ans);
return ans;
}
private void dfs(int[] nums, boolean[] used, List<Integer> path, List<List<Integer>> ans) {
if (path.size() == nums.length) {
ans.add(new ArrayList<>(path));
return;
}
for (int i = 0; i < nums.length; ++i) {
if (used[i])
continue;
if (i > 0 && nums[i] == nums[i - 1] && !used[i - 1])
continue;
used[i] = true;
path.add(nums[i]);
dfs(nums, used, path, ans);
path.remove(path.size() - 1);
used[i] = false;
}
}
}
class Solution:
def permuteUnique(self, nums: List[int]) -> List[List[int]]:
ans = []
used = [False] * len(nums)
def dfs(path: List[int]) -> None:
if len(path) == len(nums):
ans.append(path.copy())
return
for i, num in enumerate(nums):
if used[i]:
continue
if i > 0 and nums[i] == nums[i - 1] and not used[i - 1]:
continue
used[i] = True
path.append(num)
dfs(path)
path.pop()
used[i] = False
nums.sort()
dfs([])
return ans