567. Permutation in String【滑动窗口】

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].

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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;
    }
}