哈希（1）

原题：

/** * Created by gouthamvidyapradhan on 25/02/2017. * Given a string s and a non-empty string p, find all the start indices of p's anagrams in s. * <p> * Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 20,100. * <p> * The order of output does not matter. * <p> * Example 1: * <p> * Input: * s: "cbaebabacd" p: "abc" * <p> * Output: * [0, 6] * <p> * 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: * <p> * Input: * s: "abab" p: "ab" * <p> * Output: * [0, 1, 2] * <p> * 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". */

答案：

public class Anagrams {    int[] TC = new int[256];    int[] PC = new int[256];    /**     * Main method     *     * @param args     * @throws Exception     */    public static void main(String[] args) throws Exception {        List<Integer> result = new Anagrams().findAnagrams("abab", "ab");        result.forEach(System.out::print);    }    public List<Integer> findAnagrams(String s, String p) {        List<Integer> result = new ArrayList<>();        int pLen = p.length();        if (pLen > s.length()) return result;        Arrays.fill(TC, 0);        Arrays.fill(PC, 0);        for (int i = 0; i < pLen; i++) {            TC[s.charAt(i)]++;            PC[p.charAt(i)]++;        }        int i = pLen;        for (int l = s.length(); i < l; i++) {            if (compare())                result.add(i - pLen);            TC[s.charAt(i)]++;            TC[s.charAt(i - pLen)]--;        }        if (compare())            result.add(i - pLen);        return result;    }    private boolean compare() {        for (int i = 0; i < 256; i++) {            if (TC[i] != PC[i])                return false;        }        return true;    }}