leetcode-76-最小覆盖子串

题目描述：

 方法一：O(MN)

class Solution:
    def minWindow(self, s: 'str', t: 'str') -> 'str':
        from collections import Counter
        t = Counter(t)
        lookup = Counter()
        start = 0
        end = 0
        min_len = float("inf")
        res = ""
        while end < len(s):
            lookup[s[end]] += 1
            end += 1
            #print(start, end)
            while all(map(lambda x: lookup[x] >= t[x], t.keys())):
                if end - start < min_len:
                    res = s[start:end]
                    min_len = end - start
                lookup[s[start]] -= 1
                start += 1
        return res

方法二：O(M+N)

class Solution:
    def minWindow(self, s: 'str', t: 'str') -> 'str':
        from collections import defaultdict
        lookup = defaultdict(int)
        for c in t:
            lookup[c] += 1
        start = 0
        end = 0
        min_len = float("inf")
        counter = len(t)
        res = ""
        while end < len(s):
            if lookup[s[end]] > 0:
                counter -= 1
            lookup[s[end]] -= 1
            end += 1
            while counter == 0:
                if min_len > end - start:
                    min_len = end - start
                    res = s[start:end]
                if lookup[s[start]] == 0:
                    counter += 1
                lookup[s[start]] += 1
                start += 1
        return res