题目:
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], [] ]
描述:
给出一个包含重复数字的数组,要求输出该数组的所有子集
分析:
简单的搜索题,没有什么难度
代码:
class Solution {
public:
vector<vector<int>> subsetsWithDup(vector<int>& nums) {
sort(nums.begin(), nums.end());
set<vector<int>> result_set;
for (int i = 0; i <= nums.size(); ++ i) {
vector<int> current;
dfs(nums, 0, i, current, result_set);
}
vector<vector<int>> result;
result.assign(result_set.begin(), result_set.end());
return result;
}
void dfs(const vector<int>& nums, int current_index, int count, vector<int> ¤t, set<vector<int>>& result) {
if (!count) {
result.insert(current);
return;
}
if (current_index >= nums.size()) {
return;
}
dfs(nums, current_index + 1, count, current, result);
current.push_back(nums[current_index]);
dfs(nums, current_index + 1, count - 1, current, result);
current.pop_back();
}
};