[leetcode] 78. Subsets

Given a set of distinct integers, 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 = 

[1,2,3]

, a solution is:

[
[3],
[1],
[2],
[1,2,3],
[1,3],
[2,3],
[1,2],
[]
]

这道题是求集合的全部子集，题目难度为Medium。

可以用二进制数字的每一位来表示当前子集中每个元素是否存在，这样将数字从0递增到2^nums.size()，就把子集的情况遍历了一遍，所有子集就得到了。具体代码：class Solution {
public:
vector<vector<int>> subsets(vector<int>& nums) {
vector<vector<int>> rst;
sort(nums.begin(), nums.end());
int sz = nums.size();
int cnt = pow(2, sz);
for(int i=0; i<cnt; ++i) {
vector<int> curRst;
for(int j=0; j<sz; ++j) {
if(i & (1<<j)) curRst.push_back(nums[j]);
}
rst.push_back(curRst);
}
return rst;
}
};另外还可以采用回溯法，针对每个元素，依次取与不取该元素，然后通过递归得到所有子集。具体代码：

class Solution {
void getSubsets(vector<vector<int>> &rst, vector<int> &curRst, vector<int> &data, int pos) {
if(pos == data.size()) {
rst.push_back(curRst);
return;
}

curRst.push_back(data[pos++]);
getSubsets(rst, curRst, data, pos);
curRst.pop_back();

getSubsets(rst, curRst, data, pos);
}
public:
vector<vector<int>> subsets(vector<int>& nums) {
vector<vector<int>> rst;
vector<int> curRst;

sort(nums.begin(), nums.end());

getSubsets(rst, curRst, nums, 0);

return rst;
}
};

有强迫症的同学请自己实现排序。