【LeetCode】90.Subsets II

Subsets II

Given a collection of integers that might contain duplicates, nums, return all possible subsets.

Note:

• Elements in a subset must be in non-descending order.
• 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],
  []
]

这类有重复值的情况怎么处理
SubSets的算法

class Solution {
public:
    std::vector<std::vector<int> > subsets(std::vector<int> &nums) {
        std::sort(nums.begin(), nums.end());
        std::vector<std::vector<int> > res;
        std::vector<int> vec;
        subsets(res, nums, vec, 0);
        return res;
    }
private:
    void subsets(std::vector<std::vector<int> > &res, std::vector<int> &nums, std::vector<int> &vec, int begin) {
        res.push_back(vec);
        for (int i = begin; i != nums.size(); ++i) {
            vec.push_back(nums[i]);
            subsets(res, nums, vec, i + 1);
            vec.pop_back();
        }
    }
};

subsets([1,2,3,4]) = []
                     // push(1)
                     [1, subsets([2,3,4])] // if push N times in subsets([2,3,4]), the pop times is also N, so vec is also [1] after backtrack.
                     // pop(), push(2)
                     [2, subsets([3,4])]
                     // pop(), push(3)
                     [3, subsets([4])]
                     // pop(), push(4)
                     [4, subsets([])]
                     // pop()

Subsets II的解法

class Solution {
public:
    std::vector<std::vector<int> > subsetsWithDup(std::vector<int> &nums) {
        std::sort(nums.begin(), nums.end());
        std::vector<std::vector<int> > res;
        std::vector<int> vec;
        subsetsWithDup(res, nums, vec, 0);
        return res;
    }
private:
    void subsetsWithDup(std::vector<std::vector<int> > &res, std::vector<int> &nums, std::vector<int> &vec, int begin) {
        res.push_back(vec);
        for (int i = begin; i != nums.size(); ++i)
            if (i == begin || nums[i] != nums[i - 1]) { 
                vec.push_back(nums[i]);
                subsetsWithDup(res, nums, vec, i + 1);
                vec.pop_back();
            }
    }
};