178 lines
4.8 KiB
Markdown
178 lines
4.8 KiB
Markdown
## 题目地址
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https://leetcode.com/problems/permutations/description/
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## 题目描述
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```
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Given a collection of distinct integers, return all possible permutations.
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Example:
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Input: [1,2,3]
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Output:
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[
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[1,2,3],
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[1,3,2],
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[2,1,3],
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[2,3,1],
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[3,1,2],
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[3,2,1]
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]
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```
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## 思路
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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。
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这种题目其实有一个通用的解法，就是回溯法。
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网上也有大神给出了这种回溯法解题的
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[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。
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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。
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我们先来看下通用解法的解题思路，我画了一张图：
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通用写法的具体代码见下方代码区。
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## 关键点解析
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- 回溯法
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- backtrack 解题公式
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## 代码
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* 语言支持: Javascript, Python3
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Javascript Code:
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```js
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/*
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* @lc app=leetcode id=46 lang=javascript
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*
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* [46] Permutations
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*
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* https://leetcode.com/problems/permutations/description/
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*
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* algorithms
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* Medium (53.60%)
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* Total Accepted: 344.6K
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* Total Submissions: 642.9K
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* Testcase Example: '[1,2,3]'
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*
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* Given a collection of distinct integers, return all possible permutations.
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*
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* Example:
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*
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*
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* Input: [1,2,3]
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* Output:
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* [
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* [1,2,3],
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* [1,3,2],
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* [2,1,3],
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* [2,3,1],
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* [3,1,2],
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* [3,2,1]
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* ]
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*
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*
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*/
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function backtrack(list, tempList, nums) {
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if (tempList.length === nums.length) return list.push([...tempList]);
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for(let i = 0; i < nums.length; i++) {
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if (tempList.includes(nums[i])) continue;
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tempList.push(nums[i]);
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backtrack(list, tempList, nums);
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tempList.pop();
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}
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}
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/**
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* @param {number[]} nums
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* @return {number[][]}
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*/
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var permute = function(nums) {
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const list = [];
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backtrack(list, [], nums)
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return list
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};
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```
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Python3 Code:
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```Python
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class Solution:
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def permute(self, nums: List[int]) -> List[List[int]]:
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"""itertools库内置了这个函数"""
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return itertools.permutations(nums)
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def permute2(self, nums: List[int]) -> List[List[int]]:
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"""自己写回溯法"""
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res = []
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def _backtrace(nums, pre_list):
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if len(nums) <= 0:
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res.append(pre_list)
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else:
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for i in nums:
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# 注意copy一份新的调用，否则无法正常循环
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p_list = pre_list.copy()
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p_list.append(i)
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left_nums = nums.copy()
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left_nums.remove(i)
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_backtrace(left_nums, p_list)
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_backtrace(nums, [])
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return res
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```
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Python Code:
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```Python
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class Solution:
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def permute(self, nums: List[int]) -> List[List[int]]:
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"""itertools库内置了这个函数"""
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import itertools
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return itertools.permutations(nums)
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def permute2(self, nums: List[int]) -> List[List[int]]:
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"""自己写回溯法"""
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res = []
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def _backtrace(nums, pre_list):
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if len(nums) <= 0:
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res.append(pre_list)
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else:
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for i in nums:
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# 注意copy一份新的调用，否则无法正常循环
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p_list = pre_list.copy()
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p_list.append(i)
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left_nums = nums.copy()
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left_nums.remove(i)
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_backtrace(left_nums, p_list)
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_backtrace(nums, [])
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return res
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def permute3(self, nums: List[int]) -> List[List[int]]:
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"""回溯的另一种写法"""
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res = []
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length = len(nums)
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def _backtrack(start=0):
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if start == length:
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# 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
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res.append(nums[:])
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for i in range(start, length):
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nums[start], nums[i] = nums[i], nums[start]
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_backtrack(start+1)
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nums[start], nums[i] = nums[i], nums[start]
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_backtrack()
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return res
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```
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## 相关题目
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- [31.next-permutation](./31.next-permutation.md)
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- [39.combination-sum](./39.combination-sum.md)
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- [40.combination-sum-ii](./40.combination-sum-ii.md)
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- [47.permutations-ii](./47.permutations-ii.md)
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- [60.permutation-sequence](./60.permutation-sequence.md)
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- [78.subsets](./78.subsets.md)
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- [90.subsets-ii](./90.subsets-ii.md)
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- [113.path-sum-ii](./113.path-sum-ii.md)
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- [131.palindrome-partitioning](./131.palindrome-partitioning.md)
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