Find All Anagrams in a String Add to List

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”.

方法：乱序包含字母考虑哈希方式。

class Solution {
public:
vector<int> findAnagrams(string s, string p) {
vector<int> sv(26,0),pv(26,0),res;
int len = 0;
for(int i = 0 ; i < p.size(); ++i){
++pv[p[i]-'a'];
len ++;
}

for(int i = 0; i < s.size(); ++i){
++sv[s[i]-'a'];
if(i+1<len)
continue;
else{
if(sv == pv)
res.push_back(i-len+1);
--sv[s[i-len+1]-'a'];

}
}
return res;

}
};