## 题目地址
https://leetcode.com/problems/permutations/description/

## 题目描述
```
Given a collection of distinct integers, return all possible permutations.

Example:

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

```

## 思路

这道题目是求集合，并不是`求极值`，因此动态规划不是特别切合，因此我们需要考虑别的方法。

这种题目其实有一个通用的解法，就是回溯法。
网上也有大神给出了这种回溯法解题的
[通用写法](https://leetcode.com/problems/combination-sum/discuss/16502/A-general-approach-to-backtracking-questions-in-Java-(Subsets-Permutations-Combination-Sum-Palindrome-Partitioning))，这里的所有的解法使用通用方法解答。
除了这道题目还有很多其他题目可以用这种通用解法，具体的题目见后方相关题目部分。

我们先来看下通用解法的解题思路，我画了一张图：

![backtrack](../assets/problems/backtrack.png)

通用写法的具体代码见下方代码区。

## 关键点解析

- 回溯法
- backtrack 解题公式

## 代码

* 语言支持: Javascript, Python3

Javascript Code:

```js
/*
 * @lc app=leetcode id=46 lang=javascript
 *
 * [46] Permutations
 *
 * https://leetcode.com/problems/permutations/description/
 *
 * algorithms
 * Medium (53.60%)
 * Total Accepted:    344.6K
 * Total Submissions: 642.9K
 * Testcase Example:  '[1,2,3]'
 *
 * Given a collection of distinct integers, return all possible permutations.
 *
 * Example:
 *
 *
 * Input: [1,2,3]
 * Output:
 * [
 * ⁠ [1,2,3],
 * ⁠ [1,3,2],
 * ⁠ [2,1,3],
 * ⁠ [2,3,1],
 * ⁠ [3,1,2],
 * ⁠ [3,2,1]
 * ]
 *
 *
 */
function backtrack(list, tempList, nums) {
    if (tempList.length === nums.length) return list.push([...tempList]);
    for(let i = 0; i < nums.length; i++) {
        if (tempList.includes(nums[i])) continue;
        tempList.push(nums[i]);
        backtrack(list, tempList, nums);
        tempList.pop();
    }
}
/**
 * @param {number[]} nums
 * @return {number[][]}
 */
var permute = function(nums) {
    const list = [];
    backtrack(list, [], nums)
    return list
};
```
Python3 Code:
```Python
class Solution:
    def permute(self, nums: List[int]) -> List[List[int]]:
        """itertools库内置了这个函数"""
        return itertools.permutations(nums)
    
    def permute2(self, nums: List[int]) -> List[List[int]]:
        """自己写回溯法"""
        res = []
        def _backtrace(nums, pre_list):
            if len(nums) <= 0:
                res.append(pre_list)
            else:
                for i in nums:
                    # 注意copy一份新的调用，否则无法正常循环
                    p_list = pre_list.copy()
                    p_list.append(i)
                    left_nums = nums.copy()
                    left_nums.remove(i)
                    _backtrace(left_nums, p_list)
        _backtrace(nums, [])
        return res
```

Python Code:

```Python
class Solution:
    def permute(self, nums: List[int]) -> List[List[int]]:
        """itertools库内置了这个函数"""
        import itertools
        return itertools.permutations(nums)

    def permute2(self, nums: List[int]) -> List[List[int]]:
        """自己写回溯法"""
        res = []
        def _backtrace(nums, pre_list):
            if len(nums) <= 0:
                res.append(pre_list)
            else:
                for i in nums:
                    # 注意copy一份新的调用，否则无法正常循环
                    p_list = pre_list.copy()
                    p_list.append(i)
                    left_nums = nums.copy()
                    left_nums.remove(i)
                    _backtrace(left_nums, p_list)
        _backtrace(nums, [])
        return res

    def permute3(self, nums: List[int]) -> List[List[int]]:
        """回溯的另一种写法"""
        res = []
        length = len(nums)
        def _backtrack(start=0):
            if start == length:
                # nums[:] 返回 nums 的一个副本，指向新的引用，这样后续的操作不会影响已经已知解
                res.append(nums[:])
            for i in range(start, length):
                nums[start], nums[i] = nums[i], nums[start]
                _backtrack(start+1)
                nums[start], nums[i] = nums[i], nums[start]
        _backtrack()
        return res
```

## 相关题目

- [31.next-permutation](./31.next-permutation.md)
- [39.combination-sum](./39.combination-sum.md)
- [40.combination-sum-ii](./40.combination-sum-ii.md)
- [47.permutations-ii](./47.permutations-ii.md)
- [60.permutation-sequence](./60.permutation-sequence.md)(TODO)
- [78.subsets](./78.subsets.md)
- [90.subsets-ii](./90.subsets-ii.md)
- [113.path-sum-ii](./113.path-sum-ii.md)
- [131.palindrome-partitioning](./131.palindrome-partitioning.md)
