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:

1 class Solution { 2 public: 3     vector
     
      
        > subsets(vector
       
         &S) { 4         sort(S.begin(), S.end()); 5         vector
        
         
           > v(1); 6         for(int i = 0; i < S.size(); ++i) { 7             int j = v.size(); 8             while(j-- > 0) { 9                 v.push_back(v[j]);10                 v.back().push_back(S[i]);11             }12         }13         return v;14     }15 };