LeetCode 438. Find All Anagrams in a String

Given two strings s and p, return an array of all the start indices of p's anagrams in s. You may return the answer in any order.

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".

Constraints:

• 1 <= s.length, p.length <= 3 * 104
• s and p consist of lowercase English letters.

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就是要找p在s里有多少anagram，相当于是在一个字符串中找满足每个字母出现某个频率的substring（anagram -> frequency）。说是sliding window，但其实这个window的长度是固定的，也就是其实并不需要left和right，一个index遍历就行了。以及要找字母出现的频率，对于string来说可以直接用char\[\]来记录char - 'a'比用map方便多了。

代码写起来比较直观，几个自己遇到的坑：

1. 刚开始脑子没想明白，while写了半天，数组边界搞的乱七八糟。改用for以后固定只看后边界，就清楚多了。

2. 要考虑p.length() < s.length()的情况

   class Solution {
   public List findAnagrams(String s, String p) {
   List result = new ArrayList<>();
   int[] pChars = new int[26];
   int[] sChars = new int[26];
   if (s.length() < p.length()) {
   return result;
   }
   for (int i = 0; i < p.length(); i++) {
   pChars[p.charAt(i) - 'a']++;
   sChars[s.charAt(i) - 'a']++;
   }
   if (Arrays.equals(pChars, sChars)) {
   result.add(0);
   }

        for (int i = p.length(); i < s.length(); i++) {
            // remove first element in window
            sChars[s.charAt(i - p.length()) - 'a']--;
            // move window to the next
            sChars[s.charAt(i) - 'a']++;
            // update result
            if (Arrays.equals(pChars, sChars)) {
                result.add(i - p.length() + 1);
            }
        }
   
        return result;
    }

   }