Given two strings s1 and s2, write a function to return true if s2 contains the permutation of s1. In other words, one of the first string's permutations is the substring of the second string.
Example 1:
Input: s1 = "ab" s2 = "eidbaooo"
Output: True
Explanation: s2 contains one permutation of s1 ("ba").
Example 2:
Input:s1= "ab" s2 = "eidboaoo"
Output: False
Note:
The input strings only contain lower case letters.
The length of both given strings is in range [1, 10,000].
class Solution {
public boolean checkInclusion(String s1, String s2) {
if (s1 == null || s2 == null || s1.length() > s2.length()) {
return false;
}
int l1 = s1.length();
int l2 = s2.length();
int[] c1 = new int[26];
int[] c2 = new int[26];
for(char c : s1.toCharArray()){
c1[c-'a']++;
}
for(int i=0; i<l2; i++){
if(i>=l1){
c2[s2.charAt(i-l1)-'a']--;
}
c2[s2.charAt(i)-'a']++;
if(Arrays.equals(c1,c2)){
return true;
}
}
return false;
}
}