LeetCode 438. Find all letter ectopic words in the string

438. Find all alphabetic words in the string

 

【 violence 】 Count with array p The number of occurrences of each letter in , Match from beginning to end , It is equivalent to a sliding window with variable length and restriction ,O(n^2) Time complexity of , Barely make it through .

class Solution {
    public List<Integer> findAnagrams(String s, String p) {
        List<Integer> ans = new ArrayList();
        int m = s.length(), n = p.length(), l = m - n, j;
        int[] cnt = new int[26];
        for(var i = 0; i < n; i++) cnt[p.charAt(i) - 'a']++;
        for(var i = 0; i <= l; i++){
            boolean flag = true;
            int[] tmp = cnt.clone();
            for(j = 0; j < n; j++){
                char c = s.charAt(i + j);
                if(tmp[c - 'a'] == 0){
                    flag = false; break;
                }else{
                    tmp[c - 'a']--;
                }
            }
            if(flag) ans.add(i);
        }
        return ans;
    }
}

【 The sliding window 】 Fixed length sliding window , Statistics p Length of the number of characters in the window , Remove the previous characters every time you move , Plus the following characters , And p Compare the number of Chinese characters , The time complexity is reduced to O(n*26)

class Solution {

    //  The sliding window  1:28

    public List<Integer> findAnagrams(String s, String p) {
        int m = s.length(), n = p.length(), l = m - n;
        int[] cnt = new int[26], wd = new int[26];
        List<Integer> ans = new ArrayList();
        if(m < n) return ans;
        for(var i = 0; i < n; i++) cnt[p.charAt(i) - 'a']++;
        for(var i = 0; i < n; i++) wd[s.charAt(i) - 'a']++;
        if(Arrays.equals(cnt, wd)) ans.add(0);
        for(var i = n; i < m; i++){
            wd[s.charAt(i - n) - 'a']--;
            wd[s.charAt(i) - 'a']++;
            if(Arrays.equals(cnt, wd)) ans.add(i - n + 1);
        }
        return ans;
    }
}