题解 | #密码验证合格程序#
密码验证合格程序
https://www.nowcoder.com/practice/184edec193864f0985ad2684fbc86841
import java.util.*; // 注意类名必须为 Main, 不要有任何 package xxx 信息 public class Main { public static void main(String[] args) { Scanner in = new Scanner(System.in); // 注意 hasNext 和 hasNextLine 的区别 while (in.hasNext()) { // 注意 while 处理多个 case String code = in.next(); //first , we determin the length,then we go the substring part, last, we determin if there have enough type of characters. if (code.length() <= 8) { System.out.println("NG"); } else if (!subStringChecker(code)) { System.out.println("NG"); } else if (!codeChecker(code)) { System.out.println("NG"); }else{ System.out.println("OK"); } } } //判断字符数量格式是否符合要求。 public static boolean codeChecker(String part) { //We use hashset to stroe the characters that have appreared,because it helps us automaticlly repove duplicates. HashSet<String> checker = new HashSet<>(); for (int i = 0; i < part.length(); i++) { if (Character.isDigit(part.charAt(i))) { checker.add("Digit"); } else if (Character.isLowerCase(part.charAt(i))) { checker.add("LowerCase"); } else if (Character.isUpperCase(part.charAt(i))) { checker.add("UpperCase"); } else { checker.add("Other"); } } return checker.size() >= 3; } //we use sliding window to determin if there are repeated sustrings. public static boolean subStringChecker(String part) { for (int i = 0; i < part.length() - 2; i++) { //Be careful, in the code snipped"substring(i,i+n)",the index of i+n is exclude, which means that the substring's elements will be part[i],part[i+1] and part[i+2]. int indicator = part.indexOf(part.substring(i, i + 3), i + 1); if (indicator != -1) { return false; } } return true; } }