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".

class Solution {
public:
    
    int a[26],b[26];
    
    
    vector<int> findAnagrams(string s, string p) {
        int lens=s.length(),lenp=p.length(),i;
        vector<int>ans;
        if(lens==0||lens<lenp){
            return ans;
        }
        init(s,p);
        for(i=0;i<lens-lenp+1;i++){
            if(judge()){
                ans.push_back(i);
            }
            if(i!=lens-lenp){
                change(s,p,i);
            }
            
        }
        return ans;
    }
    
    void init(string s,string p){
        int i;
        for(i=0;i<p.length();i++){//数组a统计s前p.length()个字符出现的个数
            a[s[i]-'a']++;
        }
        for(i=0;i<p.length();i++){//数组b统计p中个字符出现的个数
            b[p[i]-'a']++;
        }
    }
    
    int judge(){
        for(int i=0;i<26;i++){//判断当前数组a和b中出现的字符数是否相等
            if(a[i]!=b[i]){
                return 0;
            }
        }
        return 1;
    }
    
    void change(string s,string p,int index){
        a[s[index]-'a']--;               //字符串s往后移一个，减去前面第一个字符数
        a[s[index+p.length()]-'a']++;    //字符串s往后移一个，增加新增字符次数     
    }
};