LeetCode90 Subsets II 子集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.

Example:

Input: [1,2,2]
Output:
[
  [2],
  [1],
  [1,2,2],
  [2,2],
  [1,2],
  []
]

题源： here；完整实现： here

思路：

刚开始想使用原输入nums进行迭代，但是不能处理重复的情况，所以借助另一个变量进行迭代。代码如下：

void recurse(vector<int> nums, vector<int> next, vector<vector<int>> &result){
	result.push_back(next);
	for (int i = 0; i < nums.size(); i++){
		if (i>0 && nums[i - 1] == nums[i]) continue;
		vector<int> temp = nums; vector<int> tempNext = next;
		temp.erase(temp.begin(), temp.begin() + i+1); tempNext.push_back(nums[i]);
		recurse(temp, tempNext, result);
	}
}

vector<vector<int>> subsetsWithDup(vector<int>& nums) {
	vector<vector<int>> result;
	sort(nums.begin(), nums.end());
	recurse(nums, {}, result);
	return result;
}