[46. Permutations](https://leetcode.com/problems/permutations/)

### 題目描述

Given an array `nums` of distinct integers, return *all the possible permutations.* You can return the answer in **any order**.

### 範例

**Example 1:**

Input: nums = [1,2,3]
Output: [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]

**Example 2:**

Input: nums = [0,1]
Output: [[0,1],[1,0]]

**Example 3:**

Input: nums = [1]
Output: [[1]]

**Constraints**:

* 1 <= `nums.length` <= 6
* -10 <= `nums[i]` <= 10
* All the integers of `nums` are **unique**.

### 解答
#### Python
內建作弊 by itertools

class Solution:
    def permute(self, nums: List[int]) -> List[List[int]]:
        return list(permutations(nums))

class Solution:
    def permute(self, nums: List[int]) -> List[List[int]]:
        res = []

        def backtrack(arr):
            if len(arr) == len(nums):
                res.append(arr.copy())
                return

            for num in nums:
                if num not in arr:
                    arr.append(num)
                    backtrack(arr)
                    arr.pop()

        backtrack([])
        return res

> [name=Ron Chen][time=Wed, Aug 2, 2023]

#### C++

class Solution {
public:
    vector<vector<int>> permute(vector<int>& nums) {
        vector<vector<int>> ans;
        backtrack(nums, 0, ans);
        return ans;
    }

    void backtrack(vector<int>& nums, int first, vector<vector<int>>& ans) {
        if (first >= nums.size()) {
            ans.push_back(nums);
        }
        for (int i = first; i < nums.size(); i ++) {
            swap(nums[first], nums[i]);
            backtrack(nums, first + 1, ans);
            swap(nums[first], nums[i]);
        }
    }
};

> [name=Jerry Wu][time=2 August, 2023]

### Reference

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