Given a set of distinct integers, nums, return all possible subsets.
Note:
For example,
If nums = [1,2,3], a solution is:
[ [3], [1], [2], [1,2,3], [1,3], [2,3], [1,2], [] ]
思路：遍历数组，有选择当前元素和不选当前元素两种操作，所以使用带回溯的递归
class Solution { public: vector<vector<int>> subsets(vector<int> &S) { vector<vector<int>> result; vector<int> pre; result.push_back(pre); if(S.size()==0) return result; sort(S.begin(),S.end()); DepthFirst(S,result,pre,0); return result; } void DepthFirst(vector<int> &S , vector<vector<int>> &result ,vector<int> &pre , int depth) { if (depth==S.size()) { return; } pre.push_back(S[depth]); result.push_back(pre); DepthFirst(S,result,pre,depth+1); pre.pop_back(); DepthFirst(S,result,pre,depth+1); } };
原文：http://www.cnblogs.com/qionglouyuyu/p/4855299.html