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419 lines
12 KiB
Markdown
419 lines
12 KiB
Markdown
<p align="center">
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<a href="https://programmercarl.com/other/kstar.html" target="_blank">
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<img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20210924105952.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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# 子集问题（二）
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## 90.子集II
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[力扣题目链接](https://leetcode-cn.com/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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## 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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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, vector<bool>& used) {
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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, used);
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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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## 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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### Python
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```python
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class Solution:
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def __init__(self):
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self.paths = []
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self.path = []
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def subsetsWithDup(self, nums: List[int]) -> List[List[int]]:
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nums.sort()
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self.backtracking(nums, 0)
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return self.paths
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def backtracking(self, nums: List[int], start_index: int) -> None:
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# ps.空集合仍符合要求
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self.paths.append(self.path[:])
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# Base Case
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if start_index == len(nums):
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return
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# 单层递归逻辑
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for i in range(start_index, len(nums)):
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if i > start_index and nums[i] == nums[i-1]:
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# 当前后元素值相同时，跳入下一个循环，去重
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continue
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self.path.append(nums[i])
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self.backtracking(nums, i+1)
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self.path.pop()
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```
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### Go
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```Go
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var res[][]int
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func subsetsWithDup(nums []int)[][]int {
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res=make([][]int,0)
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sort.Ints(nums)
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dfs([]int{},nums,0)
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return res
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}
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func dfs(temp, num []int, start int) {
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tmp:=make([]int,len(temp))
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copy(tmp,temp)
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res=append(res,tmp)
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for i:=start;i<len(num);i++{
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if i>start&&num[i]==num[i-1]{
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continue
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}
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temp=append(temp,num[i])
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dfs(temp,num,i+1)
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temp=temp[:len(temp)-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.slice(0))
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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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### 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
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int cmp(const void* a, const void* b) {
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return *((int*)a) - *((int*)b);
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}
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//复制函数，将当前path中的元素复制到ans中。同时记录path长度
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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 = (int**)realloc(ans, sizeof(int*) * (ansTop + 1));
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lengths[ansTop] = pathTop;
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ans[ansTop++] = tempPath;
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}
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void backTracking(int* nums, int numsSize, int startIndex, int* used) {
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//首先将当前path复制
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copy();
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//若startIndex大于数组最后一位元素的位置，返回
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if(startIndex >= numsSize)
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return ;
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int i;
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for(i = startIndex; i < numsSize; i++) {
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//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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path[pathTop++] = nums[i];
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used[i] = true;
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backTracking(nums, numsSize, i + 1, used);
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used[i] = false;
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pathTop--;
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}
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}
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int** subsetsWithDup(int* nums, int numsSize, int* returnSize, int** returnColumnSizes){
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//声明辅助变量
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path = (int*)malloc(sizeof(int) * numsSize);
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ans = (int**)malloc(0);
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lengths = (int*)malloc(sizeof(int) * 1500);
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int* used = (int*)malloc(sizeof(int) * numsSize);
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pathTop = ansTop = 0;
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//排序后查重才能生效
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qsort(nums, numsSize, sizeof(int), cmp);
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backTracking(nums, numsSize, 0, used);
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//设置一维数组和二维数组的返回大小
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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] = lengths[i];
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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 subsetsWithDup(_ nums: [Int]) -> [[Int]] {
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let nums = nums.sorted()
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var result = [[Int]]()
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var path = [Int]()
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func backtracking(startIndex: Int) {
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// 直接收集结果
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result.append(path)
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let end = nums.count
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guard startIndex < end else { return } // 终止条件
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for i in startIndex ..< end {
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if i > startIndex, nums[i] == nums[i - 1] { continue } // 跳过重复元素
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path.append(nums[i]) // 处理：收集元素
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backtracking(startIndex: i + 1) // 元素不重复访问
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path.removeLast() // 回溯
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}
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}
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backtracking(startIndex: 0)
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return result
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}
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```
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-----------------------
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<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>
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