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649 lines
18 KiB
Markdown
649 lines
18 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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# 90.子集II
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[力扣题目链接](https://leetcode.cn/problems/subsets-ii/)
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给定一个可能包含重复元素的整数数组 nums，返回该数组所有可能的子集（幂集）。
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说明：解集不能包含重复的子集。
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示例:
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* 输入: [1,2,2]
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* 输出:
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[
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[2],
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[1],
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[1,2,2],
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[2,2],
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[1,2],
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[]
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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：[回溯算法解决子集问题，如何去重？| LeetCode：90.子集II](https://www.bilibili.com/video/BV1vm4y1F71J/)，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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做本题之前一定要先做[78.子集](https://programmercarl.com/0078.子集.html)。
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这道题目和[78.子集](https://programmercarl.com/0078.子集.html)区别就是集合里有重复元素了，而且求取的子集要去重。
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那么关于回溯算法中的去重问题，**在[40.组合总和II](https://programmercarl.com/0040.组合总和II.html)中已经详细讲解过了，和本题是一个套路**。
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**剧透一下，后期要讲解的排列问题里去重也是这个套路，所以理解“树层去重”和“树枝去重”非常重要**。
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用示例中的[1, 2, 2] 来举例，如图所示： （**注意去重需要先对集合排序**）
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从图中可以看出，同一树层上重复取2 就要过滤掉，同一树枝上就可以重复取2，因为同一树枝上元素的集合才是唯一子集！
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本题就是其实就是[回溯算法：求子集问题！](https://programmercarl.com/0078.子集.html)的基础上加上了去重，去重我们在[回溯算法：求组合总和（三）](https://programmercarl.com/0040.组合总和II.html)也讲过了，所以我就直接给出代码了：
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C++代码如下：
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```CPP
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class Solution {
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private:
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vector<vector<int>> result;
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vector<int> path;
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void backtracking(vector<int>& nums, int startIndex, vector<bool>& used) {
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result.push_back(path);
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for (int i = startIndex; i < nums.size(); i++) {
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// used[i - 1] == true，说明同一树枝candidates[i - 1]使用过
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// used[i - 1] == false，说明同一树层candidates[i - 1]使用过
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// 而我们要对同一树层使用过的元素进行跳过
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if (i > 0 && nums[i] == nums[i - 1] && used[i - 1] == false) {
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continue;
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}
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path.push_back(nums[i]);
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used[i] = true;
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backtracking(nums, i + 1, used);
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used[i] = false;
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path.pop_back();
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}
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}
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public:
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vector<vector<int>> subsetsWithDup(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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sort(nums.begin(), nums.end()); // 去重需要排序
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backtracking(nums, 0, 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 * 2^n)
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* 空间复杂度: O(n)
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使用set去重的版本。
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```CPP
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class Solution {
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private:
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vector<vector<int>> result;
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vector<int> path;
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void backtracking(vector<int>& nums, int startIndex) {
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result.push_back(path);
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unordered_set<int> uset;
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for (int i = startIndex; i < nums.size(); i++) {
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if (uset.find(nums[i]) != uset.end()) {
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continue;
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}
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uset.insert(nums[i]);
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path.push_back(nums[i]);
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backtracking(nums, i + 1);
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path.pop_back();
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}
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}
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public:
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vector<vector<int>> subsetsWithDup(vector<int>& nums) {
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result.clear();
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path.clear();
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sort(nums.begin(), nums.end()); // 去重需要排序
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backtracking(nums, 0);
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return result;
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}
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};
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```
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## 补充
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本题也可以不使用used数组来去重，因为递归的时候下一个startIndex是i+1而不是0。
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如果要是全排列的话，每次要从0开始遍历，为了跳过已入栈的元素，需要使用used。
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代码如下：
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```CPP
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class Solution {
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private:
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vector<vector<int>> result;
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vector<int> path;
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void backtracking(vector<int>& nums, int startIndex) {
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result.push_back(path);
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for (int i = startIndex; i < nums.size(); i++) {
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// 而我们要对同一树层使用过的元素进行跳过
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if (i > startIndex && nums[i] == nums[i - 1] ) { // 注意这里使用i > startIndex
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continue;
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}
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path.push_back(nums[i]);
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backtracking(nums, i + 1);
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path.pop_back();
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}
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}
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public:
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vector<vector<int>> subsetsWithDup(vector<int>& nums) {
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result.clear();
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path.clear();
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sort(nums.begin(), nums.end()); // 去重需要排序
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backtracking(nums, 0);
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return result;
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}
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};
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```
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## 总结
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其实这道题目的知识点，我们之前都讲过了，如果之前讲过的子集问题和去重问题都掌握的好，这道题目应该分分钟AC。
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当然本题去重的逻辑，也可以这么写
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```cpp
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if (i > startIndex && nums[i] == nums[i - 1] ) {
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continue;
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}
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```
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## 其他语言版本
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### Java
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使用used数组
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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>> subsetsWithDup(int[] nums) {
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if (nums.length == 0){
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result.add(path);
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return result;
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}
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Arrays.sort(nums);
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used = new boolean[nums.length];
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subsetsWithDupHelper(nums, 0);
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return result;
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}
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private void subsetsWithDupHelper(int[] nums, int startIndex){
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result.add(new ArrayList<>(path));
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if (startIndex >= nums.length){
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return;
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}
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for (int i = startIndex; i < nums.length; i++){
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if (i > 0 && nums[i] == nums[i - 1] && !used[i - 1]){
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continue;
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}
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path.add(nums[i]);
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used[i] = true;
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subsetsWithDupHelper(nums, i + 1);
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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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不使用used数组
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```java
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class Solution {
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List<List<Integer>> res = new ArrayList<>();
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LinkedList<Integer> path = new LinkedList<>();
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public List<List<Integer>> subsetsWithDup( int[] nums ) {
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Arrays.sort( nums );
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subsetsWithDupHelper( nums, 0 );
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return res;
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}
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private void subsetsWithDupHelper( int[] nums, int start ) {
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res.add( new ArrayList<>( path ) );
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for ( int i = start; i < nums.length; i++ ) {
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// 跳过当前树层使用过的、相同的元素
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if ( i > start && nums[i - 1] == nums[i] ) {
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continue;
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}
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path.add( nums[i] );
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subsetsWithDupHelper( nums, i + 1 );
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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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#### Python3
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回溯 利用used数组去重
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```python
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class Solution:
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def subsetsWithDup(self, nums):
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result = []
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path = []
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used = [False] * len(nums)
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nums.sort() # 去重需要排序
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self.backtracking(nums, 0, used, path, result)
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return result
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def backtracking(self, nums, startIndex, used, path, result):
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result.append(path[:]) # 收集子集
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for i in range(startIndex, len(nums)):
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# used[i - 1] == True，说明同一树枝 nums[i - 1] 使用过
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# used[i - 1] == False，说明同一树层 nums[i - 1] 使用过
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# 而我们要对同一树层使用过的元素进行跳过
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if i > 0 and nums[i] == nums[i - 1] and not used[i - 1]:
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continue
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path.append(nums[i])
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used[i] = True
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self.backtracking(nums, i + 1, used, path, result)
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used[i] = False
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path.pop()
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```
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回溯 利用集合去重
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```python
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class Solution:
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def subsetsWithDup(self, nums):
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result = []
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path = []
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nums.sort() # 去重需要排序
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self.backtracking(nums, 0, path, result)
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return result
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def backtracking(self, nums, startIndex, path, result):
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result.append(path[:]) # 收集子集
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uset = set()
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for i in range(startIndex, len(nums)):
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if nums[i] in uset:
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continue
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uset.add(nums[i])
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path.append(nums[i])
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self.backtracking(nums, i + 1, path, result)
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path.pop()
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```
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回溯 利用递归的时候下一个startIndex是i+1而不是0去重
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```python
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class Solution:
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def subsetsWithDup(self, nums):
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result = []
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path = []
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nums.sort() # 去重需要排序
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self.backtracking(nums, 0, path, result)
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return result
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def backtracking(self, nums, startIndex, path, result):
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result.append(path[:]) # 收集子集
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for i in range(startIndex, len(nums)):
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# 而我们要对同一树层使用过的元素进行跳过
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if i > startIndex and nums[i] == nums[i - 1]:
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continue
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path.append(nums[i])
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self.backtracking(nums, i + 1, path, result)
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path.pop()
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```
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### Go
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```Go
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var (
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path []int
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res [][]int
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)
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func subsetsWithDup(nums []int) [][]int {
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path, res = make([]int, 0, len(nums)), make([][]int, 0)
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sort.Ints(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, start int) {
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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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for i := start; i < len(nums); i++ {
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if i != start && nums[i] == nums[i-1] {
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continue
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}
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path = append(path, nums[i])
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dfs(nums, i+1)
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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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### Javascript
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```Javascript
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var subsetsWithDup = function(nums) {
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let result = []
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let path = []
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let sortNums = nums.sort((a, b) => {
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return a - b
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})
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function backtracing(startIndex, sortNums) {
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result.push([...path])
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if(startIndex > nums.length - 1) {
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return
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}
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for(let i = startIndex; i < nums.length; i++) {
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if(i > startIndex && nums[i] === nums[i - 1]) {
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continue
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}
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path.push(nums[i])
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backtracing(i + 1, sortNums)
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path.pop()
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}
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}
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backtracing(0, sortNums)
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return result
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};
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```
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### TypeScript
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```typescript
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function subsetsWithDup(nums: number[]): number[][] {
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nums.sort((a, b) => a - b);
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const resArr: number[][] = [];
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backTraking(nums, 0, []);
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return resArr;
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function backTraking(nums: number[], startIndex: number, route: number[]): void {
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resArr.push([...route]);
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let length: number = nums.length;
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if (startIndex === length) return;
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for (let i = startIndex; i < length; i++) {
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if (i > startIndex && nums[i] === nums[i - 1]) continue;
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route.push(nums[i]);
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backTraking(nums, i + 1, route);
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route.pop();
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}
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}
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};
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```
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set去重版本:
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```typescript
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// 使用set去重版本
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function subsetsWithDup(nums: number[]): number[][] {
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const result: number[][] = [];
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const path: number[] = [];
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// 去重之前先排序
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nums.sort((a, b) => a - b);
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function backTracking(startIndex: number) {
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// 收集结果
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result.push([...path])
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// 此处不返回也可以因为，每次递归都会使startIndex + 1，当这个数大到nums.length的时候就不会进入递归了。
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if (startIndex === nums.length) {
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return
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}
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// 定义每一个树层的set集合
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const set: Set<number> = new Set()
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for (let i = startIndex; i < nums.length; i++) {
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// 去重
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if (set.has(nums[i])) {
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continue
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}
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set.add(nums[i])
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path.push(nums[i])
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backTracking(i + 1)
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// 回溯
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path.pop()
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}
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}
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backTracking(0)
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return result
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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>, start_index: usize, used: &mut Vec<bool>) {
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result.push(path.clone());
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let len = nums.len();
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// if start_index >= len { return; }
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for i in start_index..len {
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if i > 0 && nums[i] == nums[i - 1] && !used[i - 1] { continue; }
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path.push(nums[i]);
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used[i] = true;
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Self::backtracking(result, path, nums, i + 1, used);
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used[i] = false;
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path.pop();
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}
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}
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pub fn subsets_with_dup(mut 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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nums.sort();
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Self::backtracking(&mut result, &mut path, &nums, 0, &mut used);
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result
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}
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}
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```
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set 去重版本
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```rust
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use std::collections::HashSet;
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impl Solution {
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pub fn subsets_with_dup(mut nums: Vec<i32>) -> Vec<Vec<i32>> {
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let mut res = HashSet::new();
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let mut path = vec![];
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nums.sort();
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Self::backtracking(&nums, &mut path, &mut res, 0);
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res.into_iter().collect()
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}
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pub fn backtracking(
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nums: &Vec<i32>,
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path: &mut Vec<i32>,
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res: &mut HashSet<Vec<i32>>,
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start_index: usize,
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) {
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res.insert(path.clone());
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for i in start_index..nums.len() {
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path.push(nums[i]);
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Self::backtracking(nums, path, res, i + 1);
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path.pop();
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}
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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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//负责存放二维数组中每个数组的长度
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int* lengths;
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//快排cmp函数
|
||
int cmp(const void* a, const void* b) {
|
||
return *((int*)a) - *((int*)b);
|
||
}
|
||
|
||
//复制函数，将当前path中的元素复制到ans中。同时记录path长度
|
||
void copy() {
|
||
int* tempPath = (int*)malloc(sizeof(int) * pathTop);
|
||
int i;
|
||
for(i = 0; i < pathTop; i++) {
|
||
tempPath[i] = path[i];
|
||
}
|
||
ans = (int**)realloc(ans, sizeof(int*) * (ansTop + 1));
|
||
lengths[ansTop] = pathTop;
|
||
ans[ansTop++] = tempPath;
|
||
}
|
||
|
||
void backTracking(int* nums, int numsSize, int startIndex, int* used) {
|
||
//首先将当前path复制
|
||
copy();
|
||
//若startIndex大于数组最后一位元素的位置，返回
|
||
if(startIndex >= numsSize)
|
||
return ;
|
||
|
||
int i;
|
||
for(i = startIndex; i < numsSize; i++) {
|
||
//对同一树层使用过的元素进行跳过
|
||
if(i > 0 && nums[i] == nums[i-1] && used[i-1] == false)
|
||
continue;
|
||
path[pathTop++] = nums[i];
|
||
used[i] = true;
|
||
backTracking(nums, numsSize, i + 1, used);
|
||
used[i] = false;
|
||
pathTop--;
|
||
}
|
||
}
|
||
|
||
int** subsetsWithDup(int* nums, int numsSize, int* returnSize, int** returnColumnSizes){
|
||
//声明辅助变量
|
||
path = (int*)malloc(sizeof(int) * numsSize);
|
||
ans = (int**)malloc(0);
|
||
lengths = (int*)malloc(sizeof(int) * 1500);
|
||
int* used = (int*)malloc(sizeof(int) * numsSize);
|
||
pathTop = ansTop = 0;
|
||
|
||
//排序后查重才能生效
|
||
qsort(nums, numsSize, sizeof(int), cmp);
|
||
backTracking(nums, numsSize, 0, used);
|
||
|
||
//设置一维数组和二维数组的返回大小
|
||
*returnSize = ansTop;
|
||
*returnColumnSizes = (int*)malloc(sizeof(int) * ansTop);
|
||
int i;
|
||
for(i = 0; i < ansTop; i++) {
|
||
(*returnColumnSizes)[i] = lengths[i];
|
||
}
|
||
return ans;
|
||
}
|
||
```
|
||
|
||
### Swift
|
||
|
||
```swift
|
||
func subsetsWithDup(_ nums: [Int]) -> [[Int]] {
|
||
let nums = nums.sorted()
|
||
var result = [[Int]]()
|
||
var path = [Int]()
|
||
func backtracking(startIndex: Int) {
|
||
// 直接收集结果
|
||
result.append(path)
|
||
|
||
let end = nums.count
|
||
guard startIndex < end else { return } // 终止条件
|
||
for i in startIndex ..< end {
|
||
if i > startIndex, nums[i] == nums[i - 1] { continue } // 跳过重复元素
|
||
path.append(nums[i]) // 处理：收集元素
|
||
backtracking(startIndex: i + 1) // 元素不重复访问
|
||
path.removeLast() // 回溯
|
||
}
|
||
}
|
||
backtracking(startIndex: 0)
|
||
return result
|
||
}
|
||
```
|
||
|
||
### Scala
|
||
|
||
不使用used数组:
|
||
|
||
```scala
|
||
object Solution {
|
||
import scala.collection.mutable
|
||
def subsetsWithDup(nums: Array[Int]): List[List[Int]] = {
|
||
var result = mutable.ListBuffer[List[Int]]()
|
||
var path = mutable.ListBuffer[Int]()
|
||
var num = nums.sorted // 排序
|
||
|
||
def backtracking(startIndex: Int): Unit = {
|
||
result.append(path.toList)
|
||
if (startIndex >= num.size){
|
||
return
|
||
}
|
||
for (i <- startIndex until num.size) {
|
||
// 同一树层重复的元素不进入回溯
|
||
if (!(i > startIndex && num(i) == num(i - 1))) {
|
||
path.append(num(i))
|
||
backtracking(i + 1)
|
||
path.remove(path.size - 1)
|
||
}
|
||
}
|
||
}
|
||
|
||
backtracking(0)
|
||
result.toList
|
||
}
|
||
}
|
||
```
|
||
|
||
使用Set去重:
|
||
```scala
|
||
object Solution {
|
||
import scala.collection.mutable
|
||
def subsetsWithDup(nums: Array[Int]): List[List[Int]] = {
|
||
var result = mutable.Set[List[Int]]()
|
||
var num = nums.sorted
|
||
def backtracking(path: mutable.ListBuffer[Int], startIndex: Int): Unit = {
|
||
if (startIndex == num.length) {
|
||
result.add(path.toList)
|
||
return
|
||
}
|
||
path.append(num(startIndex))
|
||
backtracking(path, startIndex + 1) // 选择
|
||
path.remove(path.size - 1)
|
||
backtracking(path, startIndex + 1) // 不选择
|
||
}
|
||
|
||
backtracking(mutable.ListBuffer[Int](), 0)
|
||
|
||
result.toList
|
||
}
|
||
}
|
||
```
|
||
|
||
<p align="center">
|
||
<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>
|