C++回溯算法中子集问题分析探讨

class Solution {
private:
    vector<int> path;
    vector<vector<int>> result;
    void backtracking(vector<int>& nums,int index){
        result.push_back(path);
        if(index>=nums.size()) 
            return;
        for(int i=index;i<nums.size();i++){
            path.push_back(nums[i]);
            backtracking(nums,i+1);
            path.pop_back();
        }
    }
public:
    vector<vector<int>> subsets(vector<int>& nums) {
        backtracking(nums,0);
        return result;
    }
};

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

class Solution {
private:
    vector<int> path;
    vector<vector<int>> result;
    void backtracking(vector<int>& nums,int index){
        if(path.size()>1){
            result.push_back(path);
            if(index==nums.size()) return;
        }
        int used[201]={0};//记录本层元素是否重复使用，新的一层都会重新定义
        for(int i=index;i<nums.size();i++){
            if(used[nums[i]+100]==1||(!path.empty()&&nums[i]<path.back()))
                continue;
            used[nums[i]+100]=1;
            path.push_back(nums[i]);
            backtracking(nums,i+1);
            path.pop_back();
        }
    }
public:
    vector<vector<int>> findSubsequences(vector<int>& nums) {
        backtracking(nums,0);
        return result;
    }
};