【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, );
         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 + );
             vec.pop_back();
         }
     }
 };

subsets([,,,]) = []
                     // push(1)
                     [, subsets([,,])] // 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)
                     [, subsets([,])]
                     // pop(), push(3)
                     [, subsets([])]
                     // pop(), push(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, );
         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 - ]) {
                 vec.push_back(nums[i]);
                 subsetsWithDup(res, nums, vec, i + );
                 vec.pop_back();
             }
     }
 };