438. Find All Anagrams in a String 查找字符串中的所有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 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".

p都找完后，添加到结果中即可：

if (end - start == t.length()) {
                    results.add(start);
                }

class Solution {
    public List<Integer> findAnagrams(String s, String t) {
        //定义map
        HashMap<Character, Integer> map = new HashMap<>();
        List<Integer> results = new LinkedList<>();
        int start = 0, end = 0;
        
        //cc
        if (s.length() < t.length())
            return results;
        
        //定义counter
        for (char c : t.toCharArray()) {
            map.put(c, map.getOrDefault(c, 0) + 1);
        }
        int count = map.size();
        
        //while循环
        //end的减少
        while (end < s.length()) {
            char charAtEnd = s.charAt(end);
            if (map.containsKey(charAtEnd)) {
                map.put(charAtEnd, map.get(charAtEnd) - 1);
                
                if (map.get(charAtEnd) == 0)
                    count--;
            }
            end++;
            
            
            while (count == 0) {
                char charAtStart = s.charAt(start);
                if (map.containsKey(charAtStart)) {
                    map.put(charAtStart, map.get(charAtStart) + 1);
                
                if (map.get(charAtStart) > 0)
                    count++;
                }
                
                //更新head
                if (end - start == t.length()) {
                    results.add(start);
                }
                
                //start往前移
                start++;
            }
            
        }
        
        //返回
        return results;
    }
}