给定一组数(未排序), 求它们的所有组合可能.

如给定{1 2 3}, 返回:

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

算法思路:

1. 对数组排序, 从小到大;
2. 令 i = 0, 对已有组合v从后往前进行如下操作
   1. v的最后1个组合内加入第i个元素;
   2. 将新组合加入到v中

算法的理解可以通过一个例子来看:

给定S = {1 2 3},

v = [[]]

i = 0, j = 1, v = [[] [1]] // back().push_back(S[0])

i = 1, j = 2, v = [[] [1] | [1 2] [2]] // j = 2, add 2 new elems

i = 2, j = 4, v = [[] [1] [1 2] [2] | [2 3] [1 2 3] [1 3] [3]]

相同颜色前者为原有元素, 后者为增加后的元素.

代码:

 class Solution {

 public:

     vector<vector<int> > subsets(vector<int> &S) {

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

         vector<vector<int> > v();

         for(int i = ; i < S.size(); ++i) {

             int j = v.size();

             while(j-- > ) {

                 v.push_back(v[j]);

                 v.back().push_back(S[i]);

             }

         }

         return v;

     }

 };