# 647.Substring Anagrams

## 1.Description(Easy)

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 40,000.

The order of output does not matter.

**Example**

Given s =`"cbaebabacd"`p =`"abc"`

return`[0, 6]`

```
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".
```

[**Tags**](http://www.lintcode.com/en/problem/substring-anagrams/#tags)

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## 2.Code

String.substring()

```
 public List<Integer> findAnagrams(String s, String p) {

        List<Integer> result = new ArrayList<Integer>();
        if (s.length() < p.length()) {
            return result;
        }
        int len = p.length();
        for (int i = 0; i < s.length()-len+1; ++i ) {
            String current = s.substring(i,i+len);
            if (isAnagrams(current,p)){
                result.add(i);
            }
        }
        return result;
    }

    public boolean isAnagrams(String a, String b) {
        if (a.length() != b.length()) {
            return false;
        }
        int[] letter = new int[26];
        for (int i = 0; i < a.length(); ++i) {
            char currenta = a.charAt(i);
            char currentb = b.charAt(i);
            letter[currenta-'a']++;
            letter[currentb-'a']--;
        }

        for (int i = 0; i < 26; i++) {
            if (letter[i] != 0){
                return false;
            }
        }
        return true;
    }
```