请2指针→和研究，和一个哈希表 M =字符 - ＆GT;计数为包含字符串的字符不会出现在 S [l..r] 。

Keep two pointer l and r, and a hash table M = character -> count for characters in string2 that do not occur in s[l..r].

最初设定 L = 0 和研究让字符串1 [l..r ] 包含字符串2 的所有字符（如果可能的话）。你可以通过给M删除字符，直到它是空的。

Initially set l = 0 and r so that string1[l..r] contains all the characters of string2 (if possible). You do that by removing characters from M until it is empty.

然后继续通过增加研究一个在每一个步骤，然后递增→尽可能同时仍保持中号空。最低在所有的R - L + 1 （子串的长度 S [l..r] ）是溶液

Then proceed by incrementing r by one in each step and then incrementing l as much as possible while still keeping M empty. The minimum over all r - l + 1 (the length of the substring s[l..r]) is the solution.

Pythonish伪code：

Pythonish pseudocode:

n = len(string1)
M = {}   # let's say M is empty if it contains no positive values
for c in string2:
    M[c]++
l = 0
r = -1
while r + 1 < n and M not empty:
    r++
    M[string1[r]]--
if M not empty: 
    return "no solution"
answer_l, answer_r = l, r
while True:
    while M[string1[l]] < 0:
        M[string1[l]]++
        l++
    if r - l + 1 < answer_r - anwer_l + 1:
        answer_l, answer_r = l, r
    r++
    if r == n:
        break
    M[string1[r]]--
return s[answer_l..answer_r]

在为空的检查可以在O实施（1）如果在执行递增和递减操作时，保持积极的条目数

The "is empty" checks can be implemented in O(1) if you maintain the number of positive entries when performing the increment and decrement operations.

让 N 是字符串1 和 M 的长度>是长度字符串2 。

注意→和研究被永远只能增加，因此最多有O（n）的增加和因此，在大多数O（n）的指令，在过去外环执行。

Note that l and r are only ever incremented, so there are at most O(n) increments and thus at most O(n) instructions are executed in the last outer loop.

如果 M 被实现为一个数组（我假设字母表是恒定的大小），你得到的运行时 为O（N + M），这是***的。如果字母是太大，你可以使用一个哈希表来获得预期的O（N + M）。

If M is implemented as an array (I assume the alphabet is constant size), you get runtime O(n + m), which is optimal. If the alphabet is too large, you can use a hash table to get expected O(n + m).

例如执行：

string1 = "abbabcdbcb"
string2 = "cbb"

# after first loop
M = { 'a': 0, 'b': 2, 'c': 1, 'd': 0 }

# after second loop
l = 0
r = 5
M = { 'a': -2, 'b': -1, 'c': 0, 'd': 0 }

# increment l as much as possible:
l = 2
r = 5
M = { 'a': -1, 'b': 0, 'c': 0, 'd': 0 }

# increment r by one and then l as much as possible
l = 2
r = 6
M = { 'a': -1, 'b': 0, 'c': 0, 'd': -1 }

# increment r by one and then l as much as possible
l = 4
r = 7
M = { 'a': 0, 'b': 0, 'c': 0, 'd': -1 }

# increment r by one and then l as much as possible
l = 4
r = 8
M = { 'a': 0, 'b': 0, 'c': -1, 'd': -1 }

# increment r by one and then l as much as possible
l = 7
r = 9
M = { 'a': 0, 'b': 0, 'c': 0, 'd': 0 }

***的解决办法是S [7..9]。

The best solution is s[7..9].