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:
1 2 3 4 5 6 7 8 9
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:
1 2 3 4 5 6 7 8 9 10
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".
publicclassSolution{ public List<Integer> findAnagrams(String s, String p){ List<Integer> result = new ArrayList<>(); if (s == null || p.isEmpty() || s.length() < p.length()) return result;
Map<Character, Integer> ds = new HashMap<>(); Map<Character, Integer> dp = new HashMap<>();
for (int i = 0; i < p.length(); i++) { ds.put(s.charAt(i), ds.getOrDefault(s.charAt(i), 0) + 1); dp.put(p.charAt(i), dp.getOrDefault(p.charAt(i), 0) + 1); } if (isEqual(ds, dp)) { result.add(0); }
for (int i = p.length(); i < s.length(); i++) { char c = s.charAt(i); ds.put(c, ds.getOrDefault(c, 0) + 1); ds.put(s.charAt(i - p.length()), ds.getOrDefault(s.charAt(i - p.length()), 0) - 1); if (isEqual(ds, dp)) { result.add(i - p.length() + 1); } } return result; }