175 lines
4.3 KiB
Markdown
175 lines
4.3 KiB
Markdown
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## 题目地址
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https://leetcode.com/problems/subsets-ii/description/
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## 题目描述
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```
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Given a collection of integers that might contain duplicates, nums, return all possible subsets (the power set).
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Note: The solution set must not contain duplicate subsets.
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Example:
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Input: [1,2,2]
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Output:
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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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```
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## 思路
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这道题目是求集合，并不是`求极值`，因此动态规划不是特别切合，因此我们需要考虑别的方法。
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这种题目其实有一个通用的解法，就是回溯法。
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网上也有大神给出了这种回溯法解题的
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[通用写法](https://leetcode.com/problems/combination-sum/discuss/16502/A-general-approach-to-backtracking-questions-in-Java-(Subsets-Permutations-Combination-Sum-Palindrome-Partitioning))，这里的所有的解法使用通用方法解答。
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除了这道题目还有很多其他题目可以用这种通用解法，具体的题目见后方相关题目部分。
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我们先来看下通用解法的解题思路，我画了一张图：
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通用写法的具体代码见下方代码区。
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## 关键点解析
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- 回溯法
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- backtrack 解题公式
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## 代码
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* 语言支持：JS，C++，Python3
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JavaScript Code：
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```js
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/*
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* @lc app=leetcode id=90 lang=javascript
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*
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* [90] Subsets II
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*
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* https://leetcode.com/problems/subsets-ii/description/
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*
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* algorithms
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* Medium (41.53%)
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* Total Accepted: 197.1K
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* Total Submissions: 469.1K
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* Testcase Example: '[1,2,2]'
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*
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* Given a collection of integers that might contain duplicates, nums, return
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* all possible subsets (the power set).
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*
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* Note: The solution set must not contain duplicate subsets.
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*
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* Example:
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*
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*
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* Input: [1,2,2]
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* Output:
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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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*
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*
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*/
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function backtrack(list, tempList, nums, start) {
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list.push([...tempList]);
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for(let i = start; i < nums.length; i++) {
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// 和78.subsets的区别在于这道题nums可以有重复
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// 因此需要过滤这种情况
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if (i > start && nums[i] === nums[i - 1]) continue;
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tempList.push(nums[i]);
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backtrack(list, tempList, nums, i + 1)
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tempList.pop();
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}
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}
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/**
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* @param {number[]} nums
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* @return {number[][]}
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*/
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var subsetsWithDup = function(nums) {
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const list = [];
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backtrack(list, [], nums.sort((a, b) => a - b), 0, [])
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return list;
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};
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```
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C++ Code：
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```C++
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class Solution {
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private:
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void subsetsWithDup(vector<int>& nums, size_t start, vector<int>& tmp, vector<vector<int>>& res) {
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res.push_back(tmp);
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for (auto i = start; i < nums.size(); ++i) {
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if (i > start && nums[i] == nums[i - 1]) continue;
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tmp.push_back(nums[i]);
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subsetsWithDup(nums, i + 1, tmp, res);
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tmp.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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auto tmp = vector<int>();
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auto res = vector<vector<int>>();
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sort(nums.begin(), nums.end());
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subsetsWithDup(nums, 0, tmp, res);
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return res;
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}
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};
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```
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Python Code:
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```Python
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class Solution:
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def subsetsWithDup(self, nums: List[int], sorted: bool=False) -> List[List[int]]:
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"""回溯法，通过排序参数避免重复排序"""
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if not nums:
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return [[]]
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elif len(nums) == 1:
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return [[], nums]
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else:
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# 先排序，以便去重
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# 注意，这道题排序花的时间比较多
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# 因此，增加一个参数，使后续过程不用重复排序，可以大幅提高时间效率
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if not sorted:
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nums.sort()
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# 回溯法
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pre_lists = self.subsetsWithDup(nums[:-1], sorted=True)
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all_lists = [i+[nums[-1]] for i in pre_lists] + pre_lists
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# 去重
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result = []
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for i in all_lists:
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if i not in result:
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result.append(i)
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return result
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```
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## 相关题目
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- [39.combination-sum](./39.combination-sum.md)
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- [40.combination-sum-ii](./40.combination-sum-ii.md)
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- [46.permutations](./46.permutations.md)
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- [47.permutations-ii](./47.permutations-ii.md)
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- [78.subsets](./78.subsets.md)
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- [113.path-sum-ii](./113.path-sum-ii.md)
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- [131.palindrome-partitioning](./131.palindrome-partitioning.md)
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