Subsets -- LeetCode

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

Note:

• Elements in a subset must be in non-descending order.
• 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],
  []
]

思路：由于nums中每个数字都没有重复，因此该数字只有出现或不出现两种情况。

这道题有三种方法来做。第一种是用dfs来遍历每一种情况。

class Solution {
public:
    void help(vector<vector<int> >& res, vector<int>& nums, vector<int> cand, int cur)
    {
        if (cur == nums.size())
        {
            res.push_back(cand);
            return;
        }
        help(res, nums, cand, cur + 1);
        cand.push_back(nums[cur]);
        help(res, nums, cand, cur + 1);
    }
    vector<vector<int>> subsets(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        vector<vector<int> > res;
        vector<int> cand;
        help(res, nums, cand, 0);
        return res;
    }
};

第二种方法，用位运算。每一位表示nums数组中的一个数字是否出现。

class Solution {
public:
    vector<vector<int>> subsets(vector<int>& nums) {
        int tot = 1 << nums.size();
        vector<vector<int> > res;
        sort(nums.begin(), nums.end());
        for (int i = 0; i < tot; i++)
        {
            vector<int> tem;
            for (int j = 0; (1 << j) <= i; j++)
                if ((1 << j) & i)
                    tem.push_back(nums[j]);
            res.push_back(tem);
        }
        return res;
    }
};

 第三种是迭代。依次枚举每一个位置的数字，将其加到当前已有的所有子集的最后作为新的子集。

class Solution {
public:
    vector<vector<int>> subsets(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        vector<vector<int>> subs(1, vector<int>());
        for (int i = 0; i < nums.size(); i++) {
            int n = subs.size();
            for (int j = 0; j < n; j++) {
                subs.push_back(subs[j]); 
                subs.back().push_back(nums[i]);
            }
        }
        return subs;
    }
};