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".
思路: HashTable
RelatedProblems : Anagrams Two Strings Are Anagrams

public class Solution {
/**
* @param s a string
* @param p a non-empty string
* @return a list of index
*/
public List<Integer> findAnagrams(String s, String p) {
List<Integer> result = new ArrayList<>();
if (s.length() < p.length()) {
return result;
}
int[] numS = new int[256];
int[] numP = new int[256];
for (int i = 0; i < p.length(); i++) {
numS[s.charAt(i) - 'a']++;
numP[p.charAt(i) - 'a']++;
}
if (Arrays.equals(numS,numP)) {
result.add(0);
}
for (int i = p.length(); i < s.length(); i++) {
numS[s.charAt(i) - 'a']++;
numS[s.charAt(i - p.length()) - 'a']--;
if (Arrays.equals(numS,numP)) {
result.add(i - p.length() + 1);
}
}
return result;
}
}

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