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], [] ]
// OJ: https://leetcode.com/problems/subsets/
// Author: github.com/lzl124631x
// Time: O(N * 2^N)
// Space: O(N)
class Solution {
vector<vector<int>> ans;
void dfs(vector<int> &A, int i, vector<int> &s) {
if (i == A.size()) {
ans.push_back(s);
return;
}
s.push_back(A[i]); // Pick A[i]
dfs(A, i + 1, s);
s.pop_back(); // Skip A[i]
dfs(A, i + 1, s);
}
public:
vector<vector<int>> subsets(vector<int>& A) {
vector<int> s;
dfs(A, 0, s);
return ans;
}
};
// OJ: https://leetcode.com/problems/subsets
// Author: github.com/lzl124631x
// Time: O(N * 2^N)
// Space: O(N)
class Solution {
vector<vector<int>> ans;
void dfs(vector<int> &A, int start, int len, vector<int> &s) {
if (s.size() == len) {
ans.push_back(s);
return;
}
for (int i = start; i <= A.size() - len + s.size(); ++i) {
s.push_back(A[i]);
dfs(A, i + 1, len, s);
s.pop_back(); // backtrack
}
}
public:
vector<vector<int>> subsets(vector<int>& A) {
vector<int> s;
for (int len = 0; len <= A.size(); ++len) dfs(A, 0, len, s);
return ans;
}
};
// OJ: https://leetcode.com/problems/subsets
// Author: github.com/lzl124631x
// Time: O(N * 2^N)
// Space: O(1)
// Ref: https://discuss.leetcode.com/topic/2764/my-solution-using-bit-manipulation
class Solution {
public:
vector<vector<int>> subsets(vector<int>& A) {
int N = 1 << A.size();
vector<vector<int>> ans(N);
for (int i = 0; i < A.size(); ++i) { // For each numbers in A
for (int j = 0; j < N; ++j) { // check if it is in the jth subset in the output
if (j >> i & 1) ans[j].push_back(A[i]);
}
}
return ans;
}
};
Let dp[i]
be the subsets ending with A[i]
.
dp[i] = [ [k, A[i]] | k is in dp[i-1] ]
// OJ: https://leetcode.com/problems/subsets
// Author: github.com/lzl124631x
// Time: O(N * 2^N)
// Space: O(1)
class Solution {
public:
vector<vector<int>> subsets(vector<int>& A) {
vector<vector<int>> ans(1);
for (int i = 0; i < A.size(); ++i) {
int len = ans.size();
for (int j = 0; j < len; ++j) {
ans.push_back(ans[j]);
ans.back().push_back(A[i]);
}
}
return ans;
}
};