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

**解析:**Version 1,判断两个字符串是否是Anagrams,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p的字母个数并记录其长度在stat中,遍历字符串s,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,则记录索引index - n + 1到结果列表中,此时字符串继续遍历,为保证subs长度与stat长度一致,此时,subs中移除s[index - n + 1]字符,同时长度减1。
class Solution:
def findAnagrams(self, s: str, p: str) -> List[int]:
stat = collections.Counter(p)
n = len(p)
stat['length'] = n
result = []
subs = collections.defaultdict(int)
for index, ch in enumerate(s):
if ch in stat:
subs[ch] += 1
subs['length'] += 1
else:
subs = collections.defaultdict(int)
continue
if subs['length'] == stat['length']:
if stat == subs:
result.append(index - n + 1)
subs[s[index - n + 1]] -= 1
subs['length'] -= 1
return result