文章作者:Tyan 博客:noahsnail.com | CSDN | 简书

**解析:**Version 1,此题与leetcode 438非常类似,思路是一样的。判断s2是否包含s1的变换,可以采用字典的方法,即每个字母的个数及类型相等。先统计字符串s1的字母个数并记录其长度在stat中,遍历字符串s2,如果字母在stat中,则将其记录到字典subs中,否则重置subs,当subs['length'] = stat['length']时,比较二者是否相等,如果相等,直接返回True,否则,字符串继续遍历,为保证subs长度与stat长度一致,此时,subs中移除s2[index - n + 1]字符,同时长度减1。
class Solution:
def checkInclusion(self, s1: str, s2: str) -> bool:
n = len(s1)
stat = collections.Counter(s1)
stat['length'] = n
subs = collections.defaultdict(int)
for index, ch in enumerate(s2):
if ch in stat:
subs[ch] += 1
subs['length'] += 1
else:
subs = collections.defaultdict(int)
continue
if subs['length'] == stat['length']:
if stat == subs:
return True
subs[s2[index - n + 1]] -= 1
subs['length'] -= 1
return False