mirror of
https://github.com/youngyangyang04/leetcode-master.git
synced 2025-07-25 09:42:16 +08:00
495 lines
13 KiB
Markdown
495 lines
13 KiB
Markdown
<p align="center">
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<a href="https://programmercarl.com/other/xunlianying.html" target="_blank">
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<img src="../pics/训练营.png" width="1000"/>
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</a>
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<p align="center"><strong><a href="https://mp.weixin.qq.com/s/tqCxrMEU-ajQumL1i8im9A">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们收益！</strong></p>
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# 46.全排列
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[力扣题目链接](https://leetcode.cn/problems/permutations/)
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给定一个 没有重复 数字的序列，返回其所有可能的全排列。
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示例:
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* 输入: [1,2,3]
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* 输出:
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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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## 算法公开课
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**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html)：[组合与排列的区别，回溯算法求解的时候，有何不同？| LeetCode：46.全排列](https://www.bilibili.com/video/BV19v4y1S79W/)，相信结合视频再看本篇题解，更有助于大家对本题的理解**。
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## 思路
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此时我们已经学习了[77.组合问题](https://programmercarl.com/0077.组合.html)、 [131.分割回文串](https://programmercarl.com/0131.分割回文串.html)和[78.子集问题](https://programmercarl.com/0078.子集.html)，接下来看一看排列问题。
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相信这个排列问题就算是让你用for循环暴力把结果搜索出来，这个暴力也不是很好写。
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所以正如我们在[关于回溯算法，你该了解这些！](https://programmercarl.com/回溯算法理论基础.html)所讲的为什么回溯法是暴力搜索，效率这么低，还要用它？
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**因为一些问题能暴力搜出来就已经很不错了！**
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我以[1,2,3]为例，抽象成树形结构如下：
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### 回溯三部曲
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* 递归函数参数
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**首先排列是有序的，也就是说 [1,2] 和 [2,1] 是两个集合，这和之前分析的子集以及组合所不同的地方**。
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可以看出元素1在[1,2]中已经使用过了，但是在[2,1]中还要在使用一次1，所以处理排列问题就不用使用startIndex了。
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但排列问题需要一个used数组，标记已经选择的元素，如图橘黄色部分所示:
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代码如下：
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```cpp
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vector<vector<int>> result;
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vector<int> path;
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void backtracking (vector<int>& nums, vector<bool>& used)
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```
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* 递归终止条件
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可以看出叶子节点，就是收割结果的地方。
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那么什么时候，算是到达叶子节点呢？
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当收集元素的数组path的大小达到和nums数组一样大的时候，说明找到了一个全排列，也表示到达了叶子节点。
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代码如下：
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```cpp
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// 此时说明找到了一组
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if (path.size() == nums.size()) {
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result.push_back(path);
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return;
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}
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```
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* 单层搜索的逻辑
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这里和[77.组合问题](https://programmercarl.com/0077.组合.html)、[131.切割问题](https://programmercarl.com/0131.分割回文串.html)和[78.子集问题](https://programmercarl.com/0078.子集.html)最大的不同就是for循环里不用startIndex了。
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因为排列问题，每次都要从头开始搜索，例如元素1在[1,2]中已经使用过了，但是在[2,1]中还要再使用一次1。
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**而used数组，其实就是记录此时path里都有哪些元素使用了，一个排列里一个元素只能使用一次**。
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代码如下：
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```cpp
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for (int i = 0; i < nums.size(); i++) {
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if (used[i] == true) continue; // path里已经收录的元素，直接跳过
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used[i] = true;
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path.push_back(nums[i]);
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backtracking(nums, used);
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path.pop_back();
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used[i] = false;
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}
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```
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整体C++代码如下：
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```CPP
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class Solution {
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public:
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vector<vector<int>> result;
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vector<int> path;
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void backtracking (vector<int>& nums, vector<bool>& used) {
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// 此时说明找到了一组
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if (path.size() == nums.size()) {
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result.push_back(path);
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return;
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}
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for (int i = 0; i < nums.size(); i++) {
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if (used[i] == true) continue; // path里已经收录的元素，直接跳过
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used[i] = true;
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path.push_back(nums[i]);
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backtracking(nums, used);
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path.pop_back();
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used[i] = false;
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}
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}
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vector<vector<int>> permute(vector<int>& nums) {
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result.clear();
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path.clear();
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vector<bool> used(nums.size(), false);
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backtracking(nums, used);
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return result;
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}
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};
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```
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* 时间复杂度: O(n!)
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* 空间复杂度: O(n)
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## 总结
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大家此时可以感受出排列问题的不同：
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* 每层都是从0开始搜索而不是startIndex
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* 需要used数组记录path里都放了哪些元素了
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排列问题是回溯算法解决的经典题目，大家可以好好体会体会。
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## 其他语言版本
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### Java
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```java
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class Solution {
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List<List<Integer>> result = new ArrayList<>();// 存放符合条件结果的集合
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LinkedList<Integer> path = new LinkedList<>();// 用来存放符合条件结果
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boolean[] used;
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public List<List<Integer>> permute(int[] nums) {
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if (nums.length == 0){
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return result;
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}
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used = new boolean[nums.length];
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permuteHelper(nums);
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return result;
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}
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private void permuteHelper(int[] nums){
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if (path.size() == nums.length){
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result.add(new ArrayList<>(path));
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return;
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}
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for (int i = 0; i < nums.length; i++){
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if (used[i]){
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continue;
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}
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used[i] = true;
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path.add(nums[i]);
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permuteHelper(nums);
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path.removeLast();
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used[i] = false;
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}
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}
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}
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```
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```java
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// 解法2：通过判断path中是否存在数字，排除已经选择的数字
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class Solution {
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List<List<Integer>> result = new ArrayList<>();
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LinkedList<Integer> path = new LinkedList<>();
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public List<List<Integer>> permute(int[] nums) {
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if (nums.length == 0) return result;
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backtrack(nums, path);
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return result;
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}
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public void backtrack(int[] nums, LinkedList<Integer> path) {
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if (path.size() == nums.length) {
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result.add(new ArrayList<>(path));
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}
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for (int i =0; i < nums.length; i++) {
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// 如果path中已有，则跳过
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if (path.contains(nums[i])) {
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continue;
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}
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path.add(nums[i]);
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backtrack(nums, path);
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path.removeLast();
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}
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}
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}
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```
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### Python
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回溯 使用used
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```python
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class Solution:
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def permute(self, nums):
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result = []
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self.backtracking(nums, [], [False] * len(nums), result)
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return result
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def backtracking(self, nums, path, used, result):
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if len(path) == len(nums):
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result.append(path[:])
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return
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for i in range(len(nums)):
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if used[i]:
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continue
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used[i] = True
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path.append(nums[i])
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self.backtracking(nums, path, used, result)
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path.pop()
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used[i] = False
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```
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### Go
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```Go
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var (
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res [][]int
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path []int
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st []bool // state的缩写
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)
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func permute(nums []int) [][]int {
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res, path = make([][]int, 0), make([]int, 0, len(nums))
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st = make([]bool, len(nums))
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dfs(nums, 0)
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return res
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}
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func dfs(nums []int, cur int) {
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if cur == len(nums) {
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tmp := make([]int, len(path))
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copy(tmp, path)
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res = append(res, tmp)
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}
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for i := 0; i < len(nums); i++ {
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if !st[i] {
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path = append(path, nums[i])
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st[i] = true
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dfs(nums, cur + 1)
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st[i] = false
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path = path[:len(path)-1]
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}
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}
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}
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```
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### Javascript
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```js
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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 res = [], path = [];
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backtracking(nums, nums.length, []);
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return res;
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function backtracking(n, k, used) {
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if(path.length === k) {
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res.push(Array.from(path));
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return;
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}
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for (let i = 0; i < k; i++ ) {
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if(used[i]) continue;
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path.push(n[i]);
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used[i] = true; // 同支
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backtracking(n, k, used);
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path.pop();
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used[i] = false;
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}
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}
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};
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```
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### TypeScript
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```typescript
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function permute(nums: number[]): number[][] {
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const resArr: number[][] = [];
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const helperSet: Set<number> = new Set();
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backTracking(nums, []);
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return resArr;
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function backTracking(nums: number[], route: number[]): void {
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if (route.length === nums.length) {
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resArr.push([...route]);
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return;
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}
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let tempVal: number;
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for (let i = 0, length = nums.length; i < length; i++) {
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tempVal = nums[i];
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if (!helperSet.has(tempVal)) {
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route.push(tempVal);
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helperSet.add(tempVal);
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backTracking(nums, route);
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route.pop();
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helperSet.delete(tempVal);
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}
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}
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}
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};
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```
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### Rust
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```Rust
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impl Solution {
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fn backtracking(result: &mut Vec<Vec<i32>>, path: &mut Vec<i32>, nums: &Vec<i32>, used: &mut Vec<bool>) {
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let len = nums.len();
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if path.len() == len {
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result.push(path.clone());
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return;
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}
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for i in 0..len {
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if used[i] == true { continue; }
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used[i] = true;
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path.push(nums[i]);
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Self::backtracking(result, path, nums, used);
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path.pop();
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used[i] = false;
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}
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}
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pub fn permute(nums: Vec<i32>) -> Vec<Vec<i32>> {
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let mut result: Vec<Vec<i32>> = Vec::new();
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let mut path: Vec<i32> = Vec::new();
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let mut used = vec![false; nums.len()];
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Self::backtracking(&mut result, &mut path, &nums, &mut used);
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result
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}
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}
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```
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### C
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```c
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int* path;
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int pathTop;
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int** ans;
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int ansTop;
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//将used中元素都设置为0
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void initialize(int* used, int usedLength) {
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int i;
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for(i = 0; i < usedLength; i++) {
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used[i] = 0;
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}
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}
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//将path中元素拷贝到ans中
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void copy() {
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int* tempPath = (int*)malloc(sizeof(int) * pathTop);
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int i;
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for(i = 0; i < pathTop; i++) {
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tempPath[i] = path[i];
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}
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ans[ansTop++] = tempPath;
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}
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void backTracking(int* nums, int numsSize, int* used) {
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//若path中元素个数等于nums元素个数，将nums放入ans中
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if(pathTop == numsSize) {
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copy();
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return;
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}
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int i;
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for(i = 0; i < numsSize; i++) {
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//若当前下标中元素已使用过，则跳过当前元素
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if(used[i])
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continue;
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used[i] = 1;
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path[pathTop++] = nums[i];
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backTracking(nums, numsSize, used);
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//回溯
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pathTop--;
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used[i] = 0;
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}
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}
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int** permute(int* nums, int numsSize, int* returnSize, int** returnColumnSizes){
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//初始化辅助变量
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path = (int*)malloc(sizeof(int) * numsSize);
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ans = (int**)malloc(sizeof(int*) * 1000);
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int* used = (int*)malloc(sizeof(int) * numsSize);
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//将used数组中元素都置0
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initialize(used, numsSize);
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ansTop = pathTop = 0;
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backTracking(nums, numsSize, used);
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//设置path和ans数组的长度
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*returnSize = ansTop;
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*returnColumnSizes = (int*)malloc(sizeof(int) * ansTop);
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int i;
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for(i = 0; i < ansTop; i++) {
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(*returnColumnSizes)[i] = numsSize;
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}
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return ans;
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}
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```
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### Swift
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```swift
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func permute(_ nums: [Int]) -> [[Int]] {
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var result = [[Int]]()
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var path = [Int]()
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var used = [Bool](repeating: false, count: nums.count) // 记录path中已包含的元素
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func backtracking() {
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// 结束条件，收集结果
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if path.count == nums.count {
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result.append(path)
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return
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}
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for i in 0 ..< nums.count {
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if used[i] { continue } // 排除已包含的元素
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used[i] = true
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path.append(nums[i])
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backtracking()
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// 回溯
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path.removeLast()
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used[i] = false
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}
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}
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backtracking()
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return result
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}
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```
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### Scala
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```scala
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object Solution {
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import scala.collection.mutable
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def permute(nums: Array[Int]): List[List[Int]] = {
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var result = mutable.ListBuffer[List[Int]]()
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var path = mutable.ListBuffer[Int]()
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def backtracking(used: Array[Boolean]): Unit = {
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if (path.size == nums.size) {
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// 如果path的长度和nums相等，那么可以添加到结果集
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result.append(path.toList)
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return
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}
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// 添加循环守卫，只有当当前数字没有用过的情况下才进入回溯
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for (i <- nums.indices if used(i) == false) {
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used(i) = true
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path.append(nums(i))
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backtracking(used) // 回溯
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path.remove(path.size - 1)
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used(i) = false
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}
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}
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backtracking(new Array[Boolean](nums.size)) // 调用方法
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result.toList // 最终返回结果集的List形式
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}
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}
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```
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||
<p align="center">
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||
<a href="https://programmercarl.com/other/kstar.html" target="_blank">
|
||
<img src="../pics/网站星球宣传海报.jpg" width="1000"/>
|
||
</a>
|
||
|