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90.subsets-ii.md

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题目地址

https://leetcode.com/problems/subsets-ii/description/

题目描述

Given a collection of integers that might contain duplicates, nums, return all possible subsets (the power set).

Note: The solution set must not contain duplicate subsets.

Example:

Input: [1,2,2]
Output:
[
  [2],
  [1],
  [1,2,2],
  [2,2],
  [1,2],
  []
]

思路

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。

这种题目其实有一个通用的解法,就是回溯法。 网上也有大神给出了这种回溯法解题的 通用写法,这里的所有的解法使用通用方法解答。 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。

我们先来看下通用解法的解题思路,我画了一张图:

backtrack

通用写法的具体代码见下方代码区。

关键点解析

  • 回溯法
  • backtrack 解题公式

代码

  • 语言支持:JS,C++,Python3

JavaScript Code:

/*
 * @lc app=leetcode id=90 lang=javascript
 *
 * [90] Subsets II
 *
 * https://leetcode.com/problems/subsets-ii/description/
 *
 * algorithms
 * Medium (41.53%)
 * Total Accepted:    197.1K
 * Total Submissions: 469.1K
 * Testcase Example:  '[1,2,2]'
 *
 * Given a collection of integers that might contain duplicates, nums, return
 * all possible subsets (the power set).
 * 
 * Note: The solution set must not contain duplicate subsets.
 * 
 * Example:
 * 
 * 
 * Input: [1,2,2]
 * Output:
 * [
 * ⁠ [2],
 * ⁠ [1],
 * ⁠ [1,2,2],
 * ⁠ [2,2],
 * ⁠ [1,2],
 * ⁠ []
 * ]
 * 
 * 
 */
function backtrack(list, tempList, nums, start) {
    list.push([...tempList]);
    for(let i = start; i < nums.length; i++) {
        // 和78.subsets的区别在于这道题nums可以有重复
        // 因此需要过滤这种情况
        if (i > start && nums[i] === nums[i - 1]) continue;
        tempList.push(nums[i]);
        backtrack(list, tempList, nums, i + 1)
        tempList.pop();
    }
}
/**
 * @param {number[]} nums
 * @return {number[][]}
 */
var subsetsWithDup = function(nums) {
    const list = [];
    backtrack(list, [], nums.sort((a, b) => a - b), 0, [])
    return list;
};

C++ Code:

class Solution {
private:
    void subsetsWithDup(vector<int>& nums, size_t start, vector<int>& tmp, vector<vector<int>>& res) {
        res.push_back(tmp);
        for (auto i = start; i < nums.size(); ++i) {
            if (i > start && nums[i] == nums[i - 1]) continue;
            tmp.push_back(nums[i]);
            subsetsWithDup(nums, i + 1, tmp, res);
            tmp.pop_back();
        }
    }
public:
    vector<vector<int>> subsetsWithDup(vector<int>& nums) {
        auto tmp = vector<int>();
        auto res = vector<vector<int>>();
        sort(nums.begin(), nums.end());
        subsetsWithDup(nums, 0, tmp, res);
        return res;
    }
};

Python Code:

class Solution:
    def subsetsWithDup(self, nums: List[int], sorted: bool=False) -> List[List[int]]:
        """回溯法,通过排序参数避免重复排序"""
        if not nums:
            return [[]]
        elif len(nums) == 1:
            return [[], nums]
        else:
            # 先排序,以便去重
            # 注意,这道题排序花的时间比较多
            # 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
            if not sorted:
                nums.sort()
            # 回溯法
            pre_lists = self.subsetsWithDup(nums[:-1], sorted=True)
            all_lists = [i+[nums[-1]] for i in pre_lists] + pre_lists
            # 去重
            result = []
            for i in all_lists:
                if i not in result:
                    result.append(i)
            return result

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