题目要求

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.

Example 1:

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".
Example 2:

Input:
s: "abab" p: "ab"

Output:
[0, 1, 2]

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".

思路和代码

这是一个简单的双指针问题，即左指针指向可以作为起点的子数组下标，右指针则不停向右移动，将更多的元素囊括进来，从而确保该左右指针内的元素是目标数组的一个兄弟子数组（即每个字母的个数均相等）

    public List<Integer> findAnagrams(String s, String p) {
        
        List<Integer> result = new ArrayList<Integer>();
        if(p==null || s==null || p.isEmpty() || s.isEmpty()) return result;
        //左指针
        int startIndex = 0;
        //记录每个字母出现的次数
        int[] map = new int[26];
        for(char c : p.toCharArray()) {
            map[c-'a']++;
        }
        
        char[] sArray = s.toCharArray();
        //拷贝一个临时map
        int[] tmpMap = Arrays.copyOf(map, map.length);
        for(int i = 0 ; i<sArray.length ; i++) {
            int index = sArray[i] - 'a';
            //如果该字母仍有可出现次数，则说明右指针此次移动合法
            if(tmpMap[index] > 0) {
                tmpMap[index]--;
                //如果左右指针中数组的长度等于目标数组长度，说明该子数组合法，将左指针加入结果集中
                if(i - startIndex + 1 == p.length()) {
                    result.add(startIndex);
                    tmpMap[sArray[startIndex]-'a']++;
                    startIndex++;//左指针右移一位
                }
            }else if(sArray[startIndex] == sArray[i]){//如果左右指针值相等，则左指针只向右移动一个
                startIndex++;
            }else {
                //将左指针移动到当前右指针的位置上，因为中间部分的子数组一定无法构成目标数组
                startIndex = i--;
                tmpMap = Arrays.copyOf(map, map.length);
            }
        }
        return result;
    }