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LeetCode 438. Find All Anagrams in a String

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Question:

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: 

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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: 

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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".

Explanation

暴力解会超时。需要采用sliding window的方法。 参照 https://discuss.leetcode.com/topic/64434/shortest-concise-java-o-n-sliding-window-solution 另外还有一些总结的模板： https://discuss.leetcode.com/topic/30941/here-is-a-10-line-template-that-can-solve-most-substring-problems

Code

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public List<Integer> findAnagrams(String s, String p) {
    List<Integer> list = new ArrayList<>();
    if(s == null || p == null || s.length() < p.length()) return list;
    int[] chars = new int[256];
    for(Character c:p.toCharArray()) {
        chars[c] ++;
    }
    int start = 0; int end = 0; int count = p.length();
    while(end < s.length()) {
        //move right everytime, if the character exists in p's hash, decrease the count
        //current hash value >= 1 means the character is existing in p
        if (chars[s.charAt(end)] >= 1) {
            count--;
        }
        chars[s.charAt(end)]--;
        end++;
        //when the count is down to 0, means we found the right anagram
        //then add window's left to result list
        if (count == 0) {
            list.add(start);
        }
        //if we find the window's size equals to p, then we have to move left (narrow the window) to find the new match window
        //++ to reset the hash because we kicked out the left
        //only increase the count if the character is in p
        //the count >= 0 indicate it was original in the hash, cuz it won't go below 0
        if (end - start== p.length() ) {
            if (chars[s.charAt(start)] >= 0) {
                count++;
            }
            chars[s.charAt(start)]++;
            start++;
        }
    }
    return list;
}