78. Subsets

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],
[]
]

题目要求求一个集合的全部子集，并且每个子集的元素都是升序的。可以先把给定的数组排序然后用递归的方法得到全部子集。

代码：

class Solution {
public:
void backtracking(vector<int>nums,int idx,int n,vector<int>cur,vector<vector<int> >&res)
{
if(idx==n)
{
res.push_back(cur);
return;
}
backtracking(nums,idx+1,n,cur,res);
cur.push_back(nums[idx]);
backtracking(nums,idx+1,n,cur,res);
}

vector<vector<int>> subsets(vector<int>& nums) {
vector<vector<int> >res;
if(nums.empty()) return res;
int n=nums.size();
sort(nums.begin(),nums.end());
vector<int>cur;
backtracking(nums,0,n,cur,res);
return res;
}
};

也可以用非递归的方法：

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