在 java 中仅使用字符串 class 检查字谜

Question

I have an assignment to write a program that accepts 2 words and then returns whether they are an anagram of one another. Currently I am trying to loop through every letter of the first one, and check whether it matches with a letter in the second word. I will then compare the counter int with the total length of either word, which should be equal if they are anagrams.

I have an error where currently I get an infinite loop of counting upwards if the words are anagrams, and have a count to the number of letters in the words ending in the program not closing if not.

The assignment states "Use only the methods available in the String class. Do not use other Java library utilities", so I am unable to put the words into an array and use arrays.Sort.

import java.util.Scanner;

public class anagram {
  public static void main (String[] args) {
    Scanner scanner = new Scanner(System.in);
    System.out.println("Enter two strings");

    String s1 = scanner.next();
    String s2 = scanner.next();


    int counter = 0;

    int number1 = 0;
    int number2 = 0;

    if(s1.length() != s2.length()) {
        System.out.println("The strings are not anagrams");
    }   else {


        for(int i = 0; i < s1.length() ; i++) {
            for(int j = 0; j < s1.length() ; i++) {

                // number1 = i;
                // number2 = j;

                if(i == s1.length()){
                    number1 = i - 1;
                }
                if(j == s2.length()){
                    number2 = j - 1;
                }


                if(s1.charAt(number1) == s2.charAt(number2)) {
                    counter ++ ;
                    System.out.println(counter);
                }

            }

        }
        System.out.println(counter);

    }

    System.out.println(s1 + " " + s2 + " " + counter);

}

}

Answer 1

Your problem lies here

 for(int j = 0; j < s1.length() ; j++) {   ///change i to j

You are checking n^2 times, however this could be done with only 2n times.

1. Get the occurrences of all letters from 1st String (store in array)
2. Get the occurrences of all letters from 2nd String (store in array)
3. Loop through both arrays concurrently and compare whether all occurrences are the same.

Your array will have 26 elements storing the occurrences of az (if case is not sensitive)

Answer 2

In the 2nd level of your for-loop, your are incrementing i instead of j. Because j is never incremented, (j < s1.length) is always true.

import java.util.Scanner;

public class anagram {
  public static void main (String[] args) {
    Scanner scanner = new Scanner(System.in);
    System.out.println("Enter two strings");

    String s1 = scanner.next();
    String s2 = scanner.next();


    int counter = 0;

    int number1 = 0;
    int number2 = 0;

    if(s1.length() != s2.length()) {
        System.out.println("The strings are not anagrams");
    }   else {


        for(int i = 0; i < s1.length() ; i++) {
            for(int j = 0; j < s1.length() ; j++) { //LOOK HERE <======

                // number1 = i;
                // number2 = j;

                if(i == s1.length()){
                    number1 = i - 1;
                }
                if(j == s2.length()){
                    number2 = j - 1;
                }


                if(s1.charAt(number1) == s2.charAt(number2)) {
                    counter ++ ;
                    System.out.println(counter);
                }

            }

        }
        System.out.println(counter);

    }

    System.out.println(s1 + " " + s2 + " " + counter);

}
}

Answer 3

public class Main {

    static boolean isAnagram(String a, String b) {
        // Complete the function
        boolean flag = true;
        if (a.length() != b.length())
            return false;
        char str1[] = a.toCharArray();
        char str2[] = b.toCharArray();
        int countstr1[] = countLetters(str1);
        int countstr2[] = countLetters(str2);
        for (int i = 0; i < countstr1.length; i++) {
            if (countstr1[i] != countstr2[i])
                flag = false;
        }

        return flag;
    }

    public static void main(String[] args) {
        String a = "anagram";
        String b = "margana";
        boolean ret = isAnagram(a, b);
        System.out.println((ret) ? "Anagrams" : "Not Anagrams");
    }

    /** Count the occurrences of each letter */
    public static int[] countLetters(char[] chars) {
        // Declare and create an array of 26 int
        int[] counts = new int[26];
        // For each lowercase letter in the array, count it
        for (int i = 0; i < chars.length; i++)
            counts[Character.toLowerCase(chars[i]) - 'a']++;
        return counts;
    }

}

The countLetters method is the real hero in this program. This method counts and stores the frequency of each character in an array and returns the array to the calling method. This program ignores the case and determines whether the two strings are anagrams