Given a set of distinct integers, nums, return all possible subsets (the power set).
Note: The solution set must not contain duplicate subsets.
Example:
Input: nums = [1,2,3] Output: [ [3], [1], [2], [1,2,3], [1,3], [2,3], [1,2], [] ]
这道题和第77题一样,第一感觉用dfs就可以求解,这种类型的题关键点是返回条件的确定,本题和77题一样,当到达数组的末端时,达到结束条件,代码记录如下:
class Solution {
public:
vector<vector<int>> subsets(vector<int>& nums) {
vector<vector<int>> ret;
vector<int> tmp;
ret.push_back(tmp); //empty subsets
dfsSubsets(ret, tmp, nums, 0);
return ret;
}
void dfsSubsets(vector<vector<int>> &ret, vector<int> &tmp, const vector<int>& nums, int next)
{
if (next == nums.size())
return;
for (int i = next; i < nums.size(); i++)
{
tmp.push_back(nums[i]);
ret.push_back(tmp); //Attension!
dfsSubsets(ret, tmp, nums, i + 1);
tmp.pop_back();
}
}
};