There are two common methods in Java. 1. Use Arrays.sort(): convert two strings into a character array, and compare whether they are equal after sorting, and the time complexity is O(n log n); 2. Use character counting method: compare by counting the number of occurrences of each character, the time complexity is O(n), which is more efficient but slightly complicated. Both methods need to first determine whether the string length is the same, and then decide whether to handle case, space or non-alphabetical characters according to the requirements.
To determine whether two strings are anagrams, in Java, you can achieve this by comparing the sorting results of their characters. As long as the two strings have the same length and the types and quantities of characters they contain are consistent, they are all asymptotes regardless of the order.

Method 1: Use Arrays.sort()
This is the easiest and most straightforward way. Convert two strings into a character array and compare whether they are equal after sorting.
import java.util.Arrays; public class AnagramChecker { public static boolean isAnagram(String str1, String str2) { if (str1.length() != str2.length()) { return false; } char[] charArray1 = str1.toCharArray(); char[] charArray2 = str2.toCharArray(); Arrays.sort(charArray1); Arrays.sort(charArray2); return Arrays.equals(charArray1, charArray2); } }
illustrate:

- First check whether the lengths of the two strings are consistent, and if they are inconsistent, return false directly.
- Then convert the string into a character array.
- Sorts two character arrays.
- Finally, use
Arrays.equals()
to compare whether the two arrays are the same.
The time complexity of this method is about O(n log n), mainly the time spent in sorting.
Method 2: Use character counting (not sorting)
If you do not want to sort, you can use an array of size 256 to count the number of occurrences of each character (applicable to the ASCII character set).

public class AnagramChecker { public static boolean isAnagram(String str1, String str2) { if (str1.length() != str2.length()) { return false; } int[] count = new int[256]; for (char c : str1.toCharArray()) { count[c] ; } for (char c : str2.toCharArray()) { count[c]--; if (count[c] < 0) { return false; } } return true; } }
illustrate:
- Also, first determine whether the length is consistent.
- Create an array of length 256 to record the occurrences of each character.
- The first loop goes through the first string and counts each character to increase the count.
- The second loop traverses the second string, subtracting the count of the corresponding characters each time.
- If the count of a character is less than 0, it means that it is not an asymptotment.
- Finally, there is no need to check whether the entire array is 0, because the previous one has ensured that the length is consistent.
This method has a time complexity of O(n), which is more efficient, and is especially suitable for processing longer strings.
Notes and FAQs
Ignore case? If you want to ignore case, you can convert it to lowercase or uppercase before comparison:
str1 = str1.toLowerCase(); str2 = str2.toLowerCase();
Remove spaces? If the input may contain spaces and wish to ignore them:
str1 = str1.replaceAll("\\s ", ""); str2 = str2.replaceAll("\\s ", "");
Non-alphabetical characters? If you only care about letters, you can filter out other characters:
str1 = str1.replaceAll("[^a-zA-Z]", ""); str2 = str2.replaceAll("[^a-zA-Z]", "");
These preprocessing steps can be flexibly added according to specific needs.
Basically that's it. The two methods have their own applicable scenarios. The sorting method is simple and intuitive, and the character counting method is more efficient but slightly more complicated. Just choose according to actual needs.
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