2021_2_10-567_ 字符串的排列

class Solution {
     
    /**
     * frequent用于统计子字符串的每个字母出现的频率
     * account用于统计子字符串的字母种类个数
     * 当s2中的某个字符串使得account==0的时候说明存在答案
     */
    public static boolean checkInclusion(String s1, String s2) {
     
        char[] chars1 = s1.toCharArray();
        char[] chars2 = s2.toCharArray();
        int length1 = chars1.length, length2 = chars2.length;
        int[] frequent = new int[26];
        int account = 0;
        int min = Integer.MIN_VALUE;
        if (length1 > length2) {
     
            return false;
        }
        //值为min说明该字母没有出现在子字符串中
        Arrays.fill(frequent, min);
        for (char ch : chars1) {
     
            if (frequent[ch-'a'] == min) {
     
                account++;
                frequent[ch-'a'] = 0;
            }
            frequent[ch-'a']++;
        }
        for (int i = 0; i < length1; i++) {
     
            frequent[chars2[i]-'a']--;
            if (frequent[chars2[i]-'a'] == 0) {
     
                account--;
                if (account == 0) {
     
                    return true;
                }
            }
        }
        for (int i = 0; i < length2 - length1; i++) {
     
            if (frequent[chars2[i]-'a'] != min) {
     
                if (frequent[chars2[i]-'a'] == 0) {
     
                    account++;
                }
                frequent[chars2[i]-'a']++;
            }
            if (frequent[chars2[i+length1]-'a'] != min) {
     
                frequent[chars2[i+length1]-'a']--;
                if (frequent[chars2[i+length1]-'a'] == 0) {
     
                    account--;
                    if (account == 0) {
     
                        return true;
                    }
                }
            }
        }
        return false;
    }
}