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".
Solution:
class Solution {
public List<Integer> findAnagrams(String s, String p) {
Map<Character, Integer> pMap = new HashMap();
for (char c : p.toCharArray()) {
pMap.put(c, pMap.getOrDefault(c, 0) + 1);
}
int count = pMap.size();
List<Integer> result = new ArrayList();
if (s == null) return result;
char[] arr = s.toCharArray();
int left = 0;
int right = 0;
while (right < arr.length) {
char c = arr[right];
if (pMap.containsKey(c)) {
pMap.put(c, pMap.get(c) - 1);
if (pMap.get(c) == 0) {
count --;
}
}
if (right - left + 1 == p.length()) {
if (count == 0) {
result.add(left);
}
char cLeft = arr[left];
if (pMap.containsKey(cLeft)) {
pMap.put(cLeft, pMap.get(cLeft) + 1);
if (pMap.get(cLeft) == 1) {
count ++;
}
}
left ++;
}
right ++;
}
return result;
}
}