版权声明:本文为博主原创文章,未经博主允许不得转载。 https://blog.csdn.net/qq_32360995/article/details/87386820
题目:
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], [] ]
题意:
看下就懂了,这玩意叫子集还是真子集来着?
思路:
没心情写了,考砸了。。。。。。。。。难受啊,福大啊
Code:
class Solution {
public:
vector<vector<int>> subsets(vector<int>& nums) {
vector<vector<int>> q(1);
sort(nums.begin(),nums.end());
for (int i=0;i<nums.size();i++){
int size=q.size();
for (int j=0;j<size;j++){
q.push_back(q[j]);
q.back().push_back(nums[i]);
}
}
return q;
}
};