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:
求所有的子集合的问题，只不过这里的子集合里面的数字不会发生重复的，实际上比另一个会重复的还要简单一点，也是用dfs就可以解决，我还是喜欢将传递的变量生命成为private，这样方便一点，避免出现很长的参数列表：
class Solution {
public:
vector<vector<int>> subsets(vector<int>& nums) {
sort(nums.begin(), nums.end());
rawVec = nums;
tmp.clear();
ret.clear();
dfs();
return ret;
}
void dfs(int start)
{
ret.push_back(tmp);
if(start >= raw.size()) return;
if(start < rawVec.size()){
for(int i = start + ; i < rawVec.size(); ++i){
tmp.push_back(i);
dfs(i);
tmp.pop_back();
}
}
}
private:
vector<vector<int>> ret;
vector<int> tmp;
vector<int> rawVec;
};java版本的代码如下所示，基本上去除了所有的全局变量，和Subsets II的java代码基本上是相同的：
public class Solution {
public List<List<Integer>> subsets(int[] nums) {
Arrays.sort(nums);
List<List<Integer>> ret = new ArrayList<List<Integer>>();
List<Integer>tmp = new ArrayList<Integer>();
dfs(ret, tmp, 0, nums, nums.length);
return ret;
}
public void dfs(List<List<Integer>> ret, List<Integer> tmp, int start, int [] nums, int limit){
if(start > limit)
return;
else if(start == limit){
ret.add(new ArrayList(tmp));
}else{
ret.add(new ArrayList(tmp));
for(int i = start; i < limit; ++i){
tmp.add(nums[i]);
dfs(ret, tmp, i+1, nums, limit);
tmp.remove((Integer)nums[i]);
}
}
}
}