Strong Password Checker

A password is considered strong if below conditions are all met:

1. It has at least 6 characters and at most 20 characters.
2. It must contain at least one lowercase letter, at least one uppercase letter, and at least one digit.
3. It must NOT contain three repeating characters in a row ("...aaa..." is weak, but "...aa...a..." is strong, assuming other conditions are met).

Write a function strongPasswordChecker(s), that takes a string s as input, and return the MINIMUM change required to make s a strong password. If s is already strong, return 0.

Insertion, deletion or replace of any one character are all considered as one change.

思路：垃圾题，一点算法都没有。抄的答案；我看不出这里面有什么好考察的，我觉得出这个题的人水平也不怎么高。

class Solution {
   public int strongPasswordChecker(String s) {
        int one = 0, two = 0, chg = 0, p = 0, l = s.length(), r = 0, up = 0, lo = 0, d = 0;
        while (p < l) {
            char c = s.charAt(p);
            if (Character.isUpperCase(c)) up = 1;
            if (Character.isLowerCase(c)) lo = 1;
            if (Character.isDigit(c)) d = 1;
            if (p > 1 && c == s.charAt(p - 1) && c == s.charAt(p - 2)) {
                r = 2;
                while (p < l && s.charAt(p) == c) {
                    p++;
                    r++;
                }
                if (r % 3 == 0) one++;
                else if(r % 3 == 1) two++;
                chg += r / 3;
            } else p++;
        }
        int miss = 3 - up - lo - d;
        if (l < 6) {
            return Math.max(miss, 6 - l);
        } else if (l <= 20) {
            return Math.max(miss, chg);
        } else{
            int del = l - 20;
            chg -= Math.min(del, one);
            chg -= Math.min(Math.max(del - one, 0), two * 2) / 2;
            chg -= Math.max(del - one - 2 * two, 0) / 3;
            return del + Math.max(chg, miss);
        }
    }
}