78. Subsets

Leetcode Array Backtracking Bit Manipulation

题目

Given a set of distinct integers, nums, return all possible subsets (the power set).

Note: The solution set must not contain duplicate subsets.

Example:

Input: nums = [1,2,3]
Output:
[
  [3],
  [1],
  [2],
  [1,2,3],
  [1,3],
  [2,3],
  [1,2],
  []
]

分析

幂集(power set)的wikipedia页面

其实和sublists思路一样。以第$i$个元素是否出现在output中为判断条件，构建decision tree。使用backtracking。

class Solution {
public:

    void subsetsHelper(int& n, int position, vector<int>& nums, vector<int>& chosen, vector<vector<int>>& res){
        if (position==n){
            res.push_back(chosen);
            return;
        }else{
            //choose and explore
            subsetsHelper(n, position+1, nums, chosen, res); // 不选择
            int s = nums[position];
            chosen.push_back(s);
            subsetsHelper(n, position+1, nums, chosen, res); // 选择
            //unchoose
            chosen.pop_back();
        }
    }


    vector<vector<int>> subsets(vector<int>& nums) {

        vector<vector<int>> res;
        vector<int> chosen;
        int n = nums.size();
        subsetsHelper(n, 0, nums, chosen, res);
        return res;
    }
};

不过也可以使用传统的backtracking，但是其base case始终发生。

class Solution {
public:
    vector<vector<int> > subsets(vector<int> &nums) {
        vector<vector<int> > res;
        vector<int> vec;
        subsets(res, nums, nums.size(), vec, 0);
        return res;
    }
private:
    void subsets(vector<vector<int> > &res, vector<int> &nums, int n, vector<int> &vec, int begin) {
        res.push_back(vec);
        for (int i = begin; i < n; ++i) {
            vec.push_back(nums[i]);
            subsets(res, nums, n, vec, i + 1);
            vec.pop_back();
        }
    }
};