题目描述：

LeetCode 438. Find All Anagrams in a String

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".
```

Python代码：

``````class Solution(object):
def findAnagrams(self, s, p):
"""
:type s: str
:type p: str
:rtype: List[int]
"""
ls, lp = len(s), len(p)
cp = collections.Counter(p)
cs = collections.Counter()
ans = []
for i in range(ls):
cs[s[i]] += 1
if i >= lp:
cs[s[i - lp]] -= 1
if cs[s[i - lp]] == 0:
del cs[s[i - lp]]
if cs == cp:
ans.append(i - lp + 1)
return ans
``````

Python代码：

``````class Solution(object):
def findAnagrams(self, s, p):
"""
:type s: str
:type p: str
:rtype: List[int]
"""
ls, lp = len(s), len(p)
count = lp
cp = collections.Counter(p)
ans = []
for i in range(ls):
if cp[s[i]] >=1 :
count -= 1
cp[s[i]] -= 1
if i >= lp:
if cp[s[i - lp]] >= 0:
count += 1
cp[s[i - lp]] += 1
if count == 0:
ans.append(i - lp + 1)
return ans
``````

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