Find All Anagrams in a String(类似Minimum Window Substring）

Given a stringsand anon-emptystringp, find all the start indices ofp's anagrams ins.

Strings consists of lowercase English letters only and the length of both stringssandpwill 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".

分析

对比Strstr。

相同点：

s,t length要一样。

不同点：

strStr 顺序需要，所以i,j初始就是t的间隔，每次比完一起走

anagram 顺序不需要，所以Map

类似Minimum Window Substring， 一个动态的滑动窗口指针，区别在if e - ss == len(p):得结果

count遇到数字拐点0-》1 1-》0才会变化，初始值为len(t)？？？

if map: count 才++/--

map和s,e总是要++/--

class Solution:
    def findAnagrams(self, s: str, p: str) -> List[int]:
        mm = collections.defaultdict(int)
        count = 0
        for i in p:
            mm[i] +=1

        l = len(s)

        ss,e,counter =0,0,len(p)
        res = []
        while e < l : 
            if mm[s[e]] > 0:
                counter -= 1
            mm[s[e]] -= 1
            e += 1
            while counter == 0:
                if e - ss == len(p): #e已经到了下一位 所以可以直接用长度
                    res.append(ss)

                if mm[s[ss]] == 0:
                    counter += 1
                mm[s[ss]] += 1
                ss += 1
        return res