438. ectopic find all the letters in the word string
Given a string s and a non-empty string starting index p, find all s is a substring of ectopic word letter p, and returns the substring.
String contains only lowercase letters, and the character string length p and s is not more than 20,100.
Description:
Ectopic word letters refer to the same letters, but arranged in different strings.
The order of the answers that were not considered.
Example 1:
Input:
S: "cbaebabacd" P: "ABC"
Output:
[0, 6]
Explanation:
The starting index of 0 is equal to the substring "cba", which is "abc" ectopic letter words.
6 substring starting index is equal to "bac", which is "abc" ectopic letter words.
Example 2:
Input:
S: "ABAB" P: "ab &"
Output:
[0, 1, 2]
Explanation:
The starting index of 0 is equal to the substring "ab", which is "ab" ectopic letter words.
Substring starting index 1 is equal to "ba", which is "ab" ectopic letter words.
2 substring starting index is equal to "ab", which is "ab" ectopic letter words.
class Solution {
public List<Integer> findAnagrams(String s, String p) {
List<Integer> res = new ArrayList<>();
if (p.length() > s.length())
return res;
int[] pLetterCounts = new int[26];
int[] sLetterCounts = new int[26];
for (int i = 0; i < p.length(); ++i) {
pLetterCounts[p.charAt(i) - 'a']++;
sLetterCounts[s.charAt(i) - 'a']++;
}
for (int endIdx = p.length() - 1; endIdx < s.length(); ++endIdx) {
int startIdx = endIdx - p.length() + 1;
boolean isAnagrams = true;
for (int i = 0; i < 26; ++i) {
if (sLetterCounts[i] != pLetterCounts[i]) {
isAnagrams = false;
break;
}
}
if (isAnagrams)
res.add(startIdx);
if (endIdx != s.length() - 1){
sLetterCounts[s.charAt(startIdx) - 'a']--;
sLetterCounts[s.charAt(endIdx + 1) - 'a']++;
}
}
return res;
}
}