LeetCode||76. Minimum Window Substring

Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n).

For example,
S = "ADOBECODEBANC"
T = "ABC"

Minimum window is "BANC".

Note:
If there is no such window in S that covers all characters in T, return the empty string "".

If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S.

t中的字符可能为重复字符，需要先用字典记录每次字符出现的次数；要判断窗口中是否出现了所有字符，首先需要两个指针表示窗口的位置，尾指针不断往后扫，当扫到有一个窗口包含了所有T的字符，然后再收缩头指针，直到不能再收缩为止。最后记录所有可能的情况中窗口最小的。

class Solution(object):    def minWindow(self, s, t):        """        :type s: str        :type t: str        :rtype: str        """        count1 = {}        count2 = {}        for char in t:        if char not in count1:        count1[char] = 1        count2[char] = 1        else:        count1[char] += 1        count2[char] += 1        count = len(t)        start = 0        minSize = len(s) + 1        minStart = 0        for end in range(len(s)):        if s[end] in count2 and [s[end]] > 0:        count1[s[end]] -= 1        if count1[s[end]] >= 0:        count -= 1        if count == 0:        while True:        if s[start] in count2 and count2[s[start]] > 0:        if count1[s[start]] < 0:        count1[s[start]] += 1        else:        break        start += 1        if minSize > end - start + 1:        minSize = end - start + 1        minStart = start        if minSize < len(s) + 1:        return s[minStart:minStart + minSize]        else:        return ''