Leetcode | Subsets I & II

Subsets I

Given a set of distinct integers, S, 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 S =
[1,2,3], a solution is:

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

[1,2],
[]
]

Method I

我的做法是产生只有1个数的subset，然后产生有两个数的subset。这里有这么一层关系：

n+1个数的subset=前n个数的subset + 前n个数的subset都加上第n个数

比如 S=[1,2,3]，产生的序列如下：

[] ------------[]

[1]-----------[] [1]

[2] [1,2]------------[] [1] [2] [1,2]

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

为了保证所有集合是升序的，一开始需要对S进行排序。

 1 class Solution {

 2 public:

 3     vector<vector<int> > subsets(vector<int> &S) {

 4         vector<vector<int> > ret;

 5         vector<int> v;

 6         ret.push_back(v);

 7         sort(S.begin(), S.end());

 8         for(int i = 0; i < S.size(); ++i) {

 9             int n = ret.size();

10             for (int j = 0; j < n; ++j) {

11                 vector<int> nv(ret[j]);

12                 nv.push_back(S[i]);

13                 ret.push_back(nv);

14             }

15         }

16

17         return ret;

18     }

19 };

Method II

递归。每碰到一个数，要么加进去，要么不加进去。最终产生所有集合。

 1 class Solution {

 2 public:

 3     vector<vector<int> > subsets(vector<int> &S) {

 4         sort(S.begin(), S.end());

 5         vector<int> v;

 6         recursive(S, 0, v);

 7         return ret;

 8     }

 9

10     void recursive(vector<int> &S, int i, vector<int> &v) {

11         if (i == S.size()) {

12             ret.push_back(v);

13             return;

14         }

15         recursive(S, i + 1, v);

16

17         v.push_back(S[i]);

18         recursive(S, i + 1, v);

19         v.pop_back();

20     }

21

22 private:

23     vector<vector<int> > ret;

24 };