LeetCode 刷题【90. 子集 II】

墨染点香2025-09-25 1:07

90. 子集 II

自己做

解：递归选取

cpp 复制代码

class Solution {
private:
    vector<vector<int>> res;

public:
    void collect_set(vector<int>& nums, vector<int> combine, int i, bool gain){              //i为当前的下标，gain标记上个元素是否取过
        if(i == (int)nums.size()){
            res.push_back(combine);
            return;
        }

        //不取
        collect_set(nums, combine, i + 1, false);

        //取
        combine.push_back(nums[i]);
        if(i > 0 && nums[i] == nums[i - 1] && gain)                                         //相同元素且上个元素取过
            collect_set(nums, combine, i + 1, true);
        else if(i == 0 || i > 0 && nums[i] != nums[i - 1])                                  //不同元素无所谓
            collect_set(nums, combine, i + 1, true);

    }

    vector<vector<int>> subsetsWithDup(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        collect_set(nums, vector<int>(), 0, true);
        return res;
    }
};

看题解

把递归的变量掏出来

cpp 复制代码

class Solution {
private:
    vector<vector<int>> res;
    vector<int> combine;

public:
    void collect_set(vector<int>& nums, int i, bool gain){              //i为当前的下标，gain标记上个元素是否取过
        if(i == (int)nums.size()){
            res.push_back(combine);
            return;
        }

        //不取
        collect_set(nums, i + 1, false);

        //取
        if(i > 0 && nums[i] == nums[i - 1] && !gain)                                         //相同元素且上个元素没取过，直接返回
            return;

        combine.push_back(nums[i]);
        collect_set(nums, i + 1, true);
        combine.pop_back();

    }

    vector<vector<int>> subsetsWithDup(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        collect_set(nums, 0, true);
        return res;
    }
};