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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解法：

class Solution {
public:
    vector<int> findAnagrams(string s, string p) {
        unordered_map<char,int> freq;
        vector<int> result;
        if(p.size()>s.size()){
            return result;
        }
        for(int i=0;i<p.size();i++){
            freq[p[i]]++;
            freq[s[i]]--;
            if(freq[s[i]]==0){
                freq.erase(s[i]);
            }
            if(freq[p[i]]==0){
                freq.erase(p[i]);
            }
            
        }
        
        if(freq.size()==0){
            result.push_back(0);
        }
        for(int i=p.size();i<s.size();i++){
            freq[s[i]]--;
            freq[s[i-p.size()]]++;
            if(freq[s[i]]==0){
                freq.erase(s[i]);
            }
            if(freq[s[i-p.size()]]==0){
                freq.erase(s[i-p.size()]);
            }
            if(freq.size()==0){
                result.push_back(i-p.size()+1);
            }
        }
        
        return result;
    }
};