Subsets
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],
[]
]
解法一：
遍历S.size()位数的所有二进制数，1代表对应位置的元素在集合中，0代表不在。
一共2^n种情况。
详细步骤参照Subsets II
class Solution {
public:
vector<vector<int> > subsets(vector<int> &S) {
vector<vector<int> > result;
int size = S.size();
for(int i = ; i < pow(2.0, size); i ++)
{//2^size subsets
vector<int> cur;
int tag = i;
for(int j = size-; j >= ; j --)
{//for each subset, the binary presentation has size digits
if(tag% == )
cur.push_back(S[j]);
tag >>= ;
if(!tag)
break;
}
sort(cur.begin(), cur.end());
result.push_back(cur);
}
return result;
}
};解法二：
遍历所有元素，记当前元素为S[i]
遍历当前所有获得的子集，记为ret[j]
将S[i]加入ret[j]，即构成了一个新子集。
详细步骤参照Subsets II
class Solution {
public:
vector<vector<int> > subsets(vector<int> &S) {
sort(S.begin(), S.end());
vector<vector<int> > ret;
vector<int> empty;
ret.push_back(empty);
for(int i = ; i < S.size(); i ++)
{
int size = ret.size();
for(int j = ; j < size; j ++)
{
vector<int> newset = ret[j];
newset.push_back(S[i]);
ret.push_back(newset);
}
}
return ret;
}
};