给你一个字符串 s 、一个字符串 t 。返回 s 中涵盖 t 所有字符的最小子串。如果 s 中不存在涵盖 t 所有字符的子串，则返回空字符串 "" 。
注意：如果 s 中存在这样的子串，我们保证它是唯一的答案。
示例 1：
输入：s = "ADOBECODEBANC", t = "ABC"
输出："BANC"
示例 2：
输入：s = "a", t = "a"
输出："a"
提示：
1 <= s.length, t.length <= 105s 和 t 由英文字母组成进阶：你能设计一个在
o(n) 时间内解决此问题的算法吗？
## template
```python
class Solution(object):
def minWindow(self, s, t):
ls_s, ls_t = len(s), len(t)
need_to_find = [0] * 256
has_found = [0] * 256
min_begin, min_end = 0, -1
min_window = 100000000000000
for index in range(ls_t):
need_to_find[ord(t[index])] += 1
count, begin = 0, 0
for end in range(ls_s):
end_index = ord(s[end])
if need_to_find[end_index] == 0:
continue
has_found[end_index] += 1
if has_found[end_index] <= need_to_find[end_index]:
count += 1
if count == ls_t:
begin_index = ord(s[begin])
while need_to_find[begin_index] == 0 or\
has_found[begin_index] > need_to_find[begin_index]:
if has_found[begin_index] > need_to_find[begin_index]:
has_found[begin_index] -= 1
begin += 1
begin_index = ord(s[begin])
window_ls = end - begin + 1
if window_ls < min_window:
min_begin = begin
min_end = end
min_window = window_ls
if count == ls_t:
return s[min_begin: min_end + 1]
else:
return ''
if __name__ == '__main__':
s = Solution()
print (s.minWindow('a', 'a'))
```
## 答案
```python
```
## 选项
### A
```python
```
### B
```python
```
### C
```python
```