问题描述:
Given a set of distinct integers, nums, return all possible subsets (the power set).
Note: The solution set must not contain duplicate subsets.
Example:
Input: nums = [1,2,3] Output: [ [3], [1], [2], [1,2,3], [1,3], [2,3], [1,2], [] ]
思路:
这道题和前一题本质上是一样的,但是因为可以不需要进行无用的判断,所以即使递归的方法也会比较很高效。代码如下:
void enumerate(vector<int> nums, vector<int> solution, vector<vector<int>> &result){
result.push_back(solution);
for (int i = 0; i < nums.size(); i++){
vector<int> newNums = nums;
vector<int> newSolution = solution;
newNums.erase(newNums.begin(), newNums.begin() + i + 1);
newSolution.push_back(nums[i]);
enumerate(newNums, newSolution, result);
}
}
vector<vector<int>> subsets(vector<int>& nums) {
if (nums.size() == 0) return{};
vector<vector<int>> result;
sort(nums.begin(), nums.end());
enumerate(nums, {}, result);
return result;
}
贴出运行结果纪念一下: