**Difficulty: **

**Frequency: **

Given a collection of **distinct** numbers, return all possible permutations.

For example,

`[1,2,3]`

have the following permutations:

`[1,2,3]`

, `[1,3,2]`

, `[2,1,3]`

, `[2,3,1]`

, `[3,1,2]`

, and `[3,2,1]`

.

**My solution:**

**Data structure:**

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**Code:**

class Solution { public: vector<vector<int>> permute(vector<int>& nums) { vector<vector<int>> results; vector<int> tmp_vec; permute_helper(nums, results, tmp_vec); return results; } void permute_helper(vector<int> nums, vector<vector<int>> &results, vector<int> tmp_vec) { if (nums.empty()) { results.push_back(tmp_vec); return; } for (int i = 0; i < nums.size(); i++) { tmp_vec.push_back(nums[i]); vector<int> new_nums = nums; new_nums.erase(new_nums.begin() + i); permute_helper(new_nums, results, tmp_vec); tmp_vec.pop_back(); } } };

**Another solution:**

**Data structure:**

**steps:**

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**Things to learn:**

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