46. Permutations
Description
Given an array nums of distinct integers, return all the possible permutations. You can return the answer in any order .
Example 1:
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| Input: nums = [1,2,3]
Output: [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]
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Example 2:
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| Input: nums = [0,1]
Output: [[0,1],[1,0]]
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Example 3:
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| Input: nums = [1]
Output: [[1]]
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Constraints:
1 <= nums.length <= 6-10 <= nums[i] <= 10- All the integers of
nums are unique .
Hints/Notes
Solution
Language: C++
Solution with less time
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| class Solution {
public:
vector<vector<int>> res;
vector<bool> onPath;
vector<int> path;
vector<vector<int>> permute(vector<int>& nums) {
int n = nums.size();
onPath.resize(n, false);
path.resize(n, -1);
dfs(0, nums);
return res;
}
void dfs(int index, vector<int>& nums) {
if (index == path.size()) {
res.push_back(path);
return;
}
for (int i = 0; i < nums.size(); i++) {
if (!onPath[i]) {
path[index] = nums[i];
onPath[i] = true;
dfs(index + 1, nums);
onPath[i] = false;
}
}
}
};
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| class Solution {
public:
vector<vector<int>> res;
vector<vector<int>> permute(vector<int>& nums) {
unordered_set<int> available;
for (int num : nums) {
available.insert(num);
}
vector<int> cur;
dfs(nums, cur, available);
return res;
}
void dfs(vector<int>& nums, vector<int>& cur, unordered_set<int>& available) {
if (nums.size() == cur.size()) {
res.push_back(cur);
return;
}
for (int num: nums) {
if (available.contains(num)) {
available.erase(num);
cur.push_back(num);
dfs(nums, cur, available);
available.insert(num);
cur.pop_back();
}
}
}
};
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