[LeetCode] 78. Subsets 子集合

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;
    }
};

　　

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[LeetCode] Subsets II 子集合 II

 

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