leetcode题解-438. Find All Anagrams in a String

题目：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.
示例：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”.

思路：题目的关键是判断s的一个字串是否为p的anagrams。之前有道题就是起到这个作用，有兴趣的同学可以去看看。这里仍适用该思路进行判断。代码如下所示：

    public static List<Integer> findAnagrams(String s, String p) {
        List<Integer> list = new ArrayList<Integer>();
        if(s == null || p == null || s.length() < p.length())   return list;
        int lengthp = p.length();
        int lengths = s.length();
        int[] hash = new int[26];
        //将字符串p的信息保存在hash数组中
        for(char c : p.toCharArray()){
            ++hash[c - 'a'];
        }
        int left = 0, right = 0;
        while(right < lengths){
            //如果s中right处的字符在p中，则将hash数组中相应计数减一。且如果该数字大于0说明仍然属于p的anagrams范畴。将lengthp--。
            if(hash[s.charAt(right++) - 'a']-- > 0) lengthp--;
            //当lengthp==0时表明left-right中的子串属于p的anagrams，记录该索引位置
            if(lengthp == 0)    list.add(left);
            //当发现一个anagrams后，再将其信息录入到hash数组中。即相当于将p的数据再次保存。且将lengthp也恢复原来的大小
            if(right - left == p.length() && hash[s.charAt(left++) - 'a']++ >= 0)    lengthp++;
        }
        //最终返回list即可
        return list;
    }