题目地址：https://leetcode.com/problems/find-all-anagrams-in-a-string/description/open in new window
题目描述
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.
Theorder 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".
题目大意
在s中有多少个位置，从这个位置开始和p等长度的子字符串中，所包含的字符和p是一样的。
解题方法
滑动窗口
这个题考的是时间复杂度。如果判断两个切片是否是排列组合的话，时间复杂度略高，会超时。
能AC的做法是用了一个滑动窗口，每次进入窗口的字符的个数+1，超出滑动窗口的字符个数-1.
这样就一遍就搞定了，而且不用每个切片都算是不是一个排列组合。
Counter大法好，判断两个字符串是否是排列组合直接统计词频然后==判断即可。
注意如果一个词出现的次数是0，那么需要从Counter中移除，因为Counter({'a': 0, 'b': 1}) w不等于Counter({'b': 1})。
from collections import Counter
class Solution(object):
def findAnagrams(self, s, p):
"""
:type s: str
:type p: str
:rtype: List[int]
"""
answer = []
m, n = len(s), len(p)
if m < n:
return answer
pCounter = Counter(p)
sCounter = Counter(s[:n-1])
index = 0
for index in xrange(n - 1, m):
sCounter[s[index]] += 1
if sCounter == pCounter:
answer.append(index - n + 1)
sCounter[s[index - n + 1]] -= 1
if sCounter[s[index - n + 1]] == 0:
del sCounter[s[index - n + 1]]
return answer
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双指针
二刷的时候也是滑动窗口，但是使用的是双指针的解法，看上去好像没有上面这个方法更简单。
class Solution(object):
def findAnagrams(self, s, p):
"""
:type s: str
:type p: str
:rtype: List[int]
"""
count = collections.Counter()
M, N = len(s), len(p)
left, right = 0, 0
pcount = collections.Counter(p)
res = []
while right < M:
count[s[right]] += 1
if right - left + 1 == N:
if count == pcount:
res.append(left)
count[s[left]] -= 1
if count[s[left]] == 0:
del count[s[left]]
left += 1
right += 1
return res
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