LintCode 647 Substring Anagrams

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.

样例
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”.

题目的意思是给出字符串s和p，在s字符串中找出所有由p字符串颠倒字母顺序而构成的字符串，输出结果字符串的开始索引。
解一：

/**     * @param s a string     * @param p a non-empty string     * @return a list of index     */    public List<Integer> findAnagrams(String s, String p) {        ArrayList<Integer> result = new ArrayList<>();        if (s == null || p == null) {            return result;        }        int left = 0, right = 0, count = p.length();        int map[] = new int[256];        char[] sc = s.toCharArray();        for (char c : p.toCharArray()) {            map[c]++;        }        while (right < s.length()) {            if (map[sc[right++]]-- >= 1) {                count--;            }            if (count == 0) {                result.add(left);            }            if (right - left == p.length() && map[sc[left++]]++ >= 0) {                count++;            }        }        return result;    }