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".

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【思路】暴力hash  遍历匹配 hash table
class Solution {
public:
vector<int> findAnagrams(string s, string p) {
vector<int>hash1(256);
vector<int>hash2(256);
for(auto ch : p) hash1[ch]++;
int lenp=p.size();
int lens=s.size();
vector<int> ans;
for(int i =0;i<lens;i++){
hash2[s[i]]++;
if(i>=lenp) hash2[s[i-lenp]]--;
if(hash1 ==hash2 ) ans.push_back(i-lenp+1);
}
return ans;
}
};