1. Leetcode 78. Subsets
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], [] ]
Difficulty:Medium
Category:
2. Analyze
求子集集合的问题,我们可以每次处理一个元素,将当前的元素加到之前存在的所有子集里面,这样能够很方便的处理。
需要提前压进去一个空集
3. Solution
3.1. Solution 1: DFS
// Runtime: 8 ms, faster than 100.00% of C++ online submissions for Subsets.
// Memory Usage: 9 MB, less than 100.00% of C++ online submissions for Subsets.
class Solution {
public:
vector<vector<int>> subsets(vector<int>& nums) {
if (nums.empty()) return {{}};
vector<vector<int>> ans;
vector<int> out;
ans.emplace_back(out);
subsetsDFS(nums, 0, out, ans);
return ans;
}
private:
void subsetsDFS(vector<int>& nums, int index, vector<int>& out, vector<vector<int>>& ans) {
if (index == nums.size()) return;
for (int i = index; i < nums.size(); ++i) {
out.emplace_back(nums[i]);
ans.emplace_back(out);
subsetsDFS(nums, i + 1, out, ans);
out.pop_back();
}
}
};
3.2. Solution 2: BFS
// Non-recursion
class Solution {
public:
vector<vector<int> > subsets(vector<int> &S) {
vector<vector<int> > res(1);
sort(S.begin(), S.end());
for (int i = 0; i < S.size(); ++i) {
int len = res.size();
for (int j = 0; j < len; ++j) {
res.push_back(res[j]);
res.back().push_back(S[i]);
}
}
return res;
}
};
4. Updated
- 03/01/2019 Review(BFS: 5mins)