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。

 1 class Solution {
 2 public:
 3     void help(vector<vector<int> >& res, vector<int>& nums, vector<int> cand, int cur)
 4     {
 5         if (cur == nums.size())
 6         {
 7             res.push_back(cand);
 8             return;
 9         }
10         help(res, nums, cand, cur + 1);
11         cand.push_back(nums[cur]);
12         help(res, nums, cand, cur + 1);
13     }
14     vector<vector<int>> subsets(vector<int>& nums) {
15         sort(nums.begin(), nums.end());
16         vector<vector<int> > res;
17         vector<int> cand;
18         help(res, nums, cand, 0);
19         return res;
20     }
21 };

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。

 1 class Solution {
 2 public:
 3     vector<vector<int>> subsets(vector<int>& nums) {
 4         int tot = 1 << nums.size();
 5         vector<vector<int> > res;
 6         sort(nums.begin(), nums.end());
 7         for (int i = 0; i < tot; i++)
 8         {
 9             vector<int> tem;
10             for (int j = 0; (1 << j) <= i; j++)
11                 if ((1 << j) & i)
12                     tem.push_back(nums[j]);
13             res.push_back(tem);
14         }
15         return res;
16     }
17 };

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。

 1 class Solution {
 2 public:
 3     vector<vector<int>> subsets(vector<int>& nums) {
 4         sort(nums.begin(), nums.end());
 5         vector<vector<int>> subs(1, vector<int>());
 6         for (int i = 0; i < nums.size(); i++) {
 7             int n = subs.size();
 8             for (int j = 0; j < n; j++) {
 9                 subs.push_back(subs[j]); 
10                 subs.back().push_back(nums[i]);
11             }
12         }
13         return subs;
14     }
15 };