[LeetCode] 90. Subsets II 子集合 II

Given a collection of integers that might contain duplicates, nums, return all possible subsets (the power set).

Note: The solution set must not contain duplicate subsets.

For example,
If nums = [1,2,2], a solution is:

[
  [2],
  [1],
  [1,2,2],
  [2,2],
  [1,2],
  []
]

78. Subsets 的拓展，这题数组里面可能含有重复元素。

Python：

class Solution(object):
    def subsetsWithDup(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        nums.sort()
        result = [[]]
        previous_size = 0
        for i in xrange(len(nums)):
            size = len(result)
            for j in xrange(size):
                # Only union non-duplicate element or new union set.
                if i == 0 or nums[i] != nums[i - 1] or j >= previous_size:
                    result.append(list(result[j]))
                    result[-1].append(nums[i])
            previous_size = size
        return result

Python:

class Solution2(object):
    def subsetsWithDup(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        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])
            if cur not in result:
                result.append(cur)
            i += 1
            
        return result

Python:　　

class Solution3(object):
    def subsetsWithDup(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        result = []
        self.subsetsWithDupRecu(result, [], sorted(nums))
        return result
    
    def subsetsWithDupRecu(self, result, cur, nums):
        if not nums:
            if cur not in result:
                result.append(cur)
        else:
            self.subsetsWithDupRecu(result, cur, nums[1:])
            self.subsetsWithDupRecu(result, cur + [nums[0]], nums[1:])　　

C++：

class Solution {
public:
    vector<vector<int>> subsetsWithDup(vector<int> &nums) {
        vector<vector<int>> result(1);
        sort(nums.begin(), nums.end());
        size_t previous_size = 0;
        for (size_t i = 0; i < nums.size(); ++i) {
            const size_t size = result.size();
            for (size_t j = 0; j < size; ++j) {
                // Only union non-duplicate element or new union set.
                if (i == 0 || nums[i] != nums[i - 1] || j >= previous_size) {
                    result.emplace_back(result[j]);
                    result.back().emplace_back(nums[i]);
                }
            }
            previous_size = size;
        }
        return result;
    }
};

　　

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[LeetCode] 78. Subsets 子集合

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