78. Subsets(backtracking 套路)

Given a set of distinct integers, nums, return all possible subsets.

Note: The solution set must not contain duplicate subsets.

For example,
If nums = [1,2,3], a solution is:

[
[3],
[1],
[2],
[1,2,3],
[1,3],
[2,3],
[1,2],
[]
]

解答：这道题不算难，还是套路，但是我的做法比较慢，别的人用位运算更快，不过我没仔细看别人怎么做的。应该注意，这道题里面的套路也就是backtracking的套路基本就是这道题的答案，可以记一下：

class Solution {
public:
    vector<vector<int>> subsets(vector<int>& nums) {
          vector<vector<int>> result;
          vector<int> res;
          sub(nums,result,res,0,nums.size());
          return result;
    }
    void sub(vector<int>& nums,vector<vector<int>>& result, vector<int> res,int begin,int target)
    {
          result.push_back(res);
          for(int i=begin;i1,target);
              res.pop_back();
          }

    }


};

别人的位操作做法：

class Solution {
public:
    vector<vector<int>> subsets(vector<int>& nums) {
          vector<vector<int>> result;
          vector<int> res;
          sub(nums,result,res,0,nums.size());
          return result;
    }
    void sub(vector<int>& nums,vector<vector<int>>& result, vector<int> res,int begin,int target)
    {
          result.push_back(res);
          for(int i=begin;i1,target);
              res.pop_back();
          }

    }


};