Given a set of distinct integers, nums, return all possible subsets (the power set).
Note: 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], [] ]
Python:
class Solution(object):
def subsets(self, nums):
"""
:type nums: List[int]
:rtype: List[List[int]]
"""
nums.sort()
result = [[]]
for i in xrange(len(nums)):
size = len(result)
for j in xrange(size):
result.append(list(result[j]))
result[-1].append(nums[i])
return result
Python:
class Solution2(object):
def subsets(self, nums):
result = []
i, count = 0, 1 << len(nums)
nums.sort()
while i < count:
cur = []
for j in xrange(len(nums)):
if i & 1 << j:
cur.append(nums[j])
result.append(cur)
i += 1
return result
Python:
class Solution3(object):
def subsets(self, nums):
"""
:type nums: List[int]
:rtype: List[List[int]]
"""
return self.subsetsRecu([], sorted(nums))
def subsetsRecu(self, cur, nums):
if not nums:
return [cur]
return self.subsetsRecu(cur, nums[1:]) + self.subsetsRecu(cur + [nums[0]], nums[1:])
C++:
class Solution {
public:
vector<vector<int> > subsets(vector<int> &nums) {
vector<vector<int>> result(1);
sort(nums.begin(), nums.end());
for (size_t i = 0; i < nums.size(); ++i) {
const size_t size = result.size();
for (size_t j = 0; j < size; ++j) {
result.emplace_back(result[j]);
result.back().emplace_back(nums[i]);
}
}
return result;
}
};
类似题目：
[LeetCode] Subsets II 子集合 II
