Given a string s and a non-empty string p, find all the start indices of p’s anagrams in s.
Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 20,100.
The order of output does not matter.
Example 1:
Input:
s: "cbaebabacd" p: "abc"
Output:
[0, 6]
Explanation:
The substring with start index = 0 is "cba", which is an anagram of "abc".
The substring with start index = 6 is "bac", which is an anagram of "abc".
Example 2:
Input:
s: "abab" p: "ab"
Output:
[0, 1, 2]
Explanation:
The substring with start index = 0 is "ab", which is an anagram of "ab".
The substring with start index = 1 is "ba", which is an anagram of "ab".
The substring with start index = 2 is "ab", which is an anagram of "ab".
Solution
先计算一下p的长度， 然后按顺序从s中取长度和p等长的字符串，看看他们是不是anagram, 我们可以排序这两个字符串，然后看看是不是相等，如果相等则把索引放进最终的结果中，但这样会超时，我们用来替换。然后移到下一个字符，把新字符放在里去，同时把第一个从Counter里减少一个,如果为0，则移除。
from collections import Counter
class Solution:
def findAnagrams(self, s: str, p: str) -> List[int]:
res =[]
cp = Counter(p)
lp =len(p)
i = 0
sc= Counter(s[i:i+ lp-1])
while (i+lp<=len(s)):
sc[s[i+lp -1]] +=1
if sc ==cp:
res.append(i)
sc[s[i]] -=1
if sc[s[i]] ==0:
del sc[s[i]]
i = i+1
return res
