我觉得可以，但是超时了
class Solution:
def findAnagrams(self, s: str, p: str) -> List[int]:
s = list(s)
p = list(p)
l = len(p)
res = []
for i in range(len(s)):
m = 0
k = s[i:i+l]
for j in range(l):
if p[j] in k :
k.remove(p[j])
else:
m += 1
if m == 0:
res.append(i)
return res
官方答案(滑动窗口)
class Solution:
def findAnagrams(self, s: str, p: str) -> List[int]:
slen, plen = len(s), len(p)
need = Counter(p)
window = Counter()
left, right, valid = 0, 0, 0
res = []
while right < slen:
c = s[right]
right += 1
if need.get(c):
window[c] += 1
if window[c] == need[c]:
valid += 1
while right - left >= plen:
if valid == len(need):
res.append(left)
d = s[left]
left += 1
if need.get(d):
if window[d] == need[d]:
valid -= 1
window[d] -= 1
return res
