【LeetCode】78. Subsets (2 solutions)

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 = 0; i < pow(2.0, size); i ++)
{//2^size subsets
vector<int> cur;
int tag = i;
for(int j = size-1; j >= 0; j --)
{//for each subset, the binary presentation has size digits
if(tag%2 == 1)
cur.push_back(S[j]);
tag >>= 1;
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 = 0; i < S.size(); i ++)
{
int size = ret.size();
for(int j = 0; j < size; j ++)
{
vector<int> newset = ret[j];
newset.push_back(S[i]);
ret.push_back(newset);
}
}
return ret;
}
};