[LeetCode系列]子集枚举问题[无重复元素]

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<vector<int> > subsets(vector<int> &S) {
 4         sort(S.begin(), S.end());
 5         vector<vector<int> > 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 };