Java Anagram-它是除排序之外最简单的解决方案

Question

I am creating the algorithm to check if two given words are Anagram or not. I came up with this simple solution. But I think it's too simple and I can not find this much simpler solution. Am I right with this solution? or something wrong with this solution?

public boolean checkAnagram(String a, String b){
    if(a.length() != b.length()) return false;
    a=a.toLowerCase();
    b=b.toLowerCase();
    int length= a.length();
    int sum1=0;
    int sum2=0;
    for(int i=0;i<length;i++){
        sum1 += a.charAt(i);
        sum2 +=b.charAt(i);
    }

    if(sum1==sum2){
        return true;
    } else {
        return false;
    }
}

Answer 1

This solution is wrong - summing values loses a lot of data, specifically, how the values are distributed. Eg, if you feed "AC" and "BB" to this function, you'll get true , which is the wrong result.

Instead, you should count the number of times each character appears in each string and compare them:

public static boolean checkAnagram(String a, String b){
    // optimiztion, will also work without it
    if (a.length() != b.length()) {
        return false;
    }
    Map<Character, Long> aMap = countChars(a);
    Map<Character, Long> bMap = countChars(b);
    return a.equals(b);
}

private static Map<Character, Long> countChars(String s) {
    return s.chars()
            .mapToObj(c -> Character.toLowerCase((char) c))
            .collect(Collectors.groupingBy(Function.identity(), 
                     Collectors.counting()));
}

Answer 2

This won't work as ac is equal to bb .

You need to actually keep track of the characters:

int[] aArr = new int[26];
int[] bArr = new int[26];
for(char c : a. toLowerCase().toCharArray()) {
    aArr[c - 'a']++;
}
// same for b
return Arrays.equals(aArr, bArr);

Time complexity is O(n) (with the length check)

Answer 3

convert the strings to char arrays and use the Arrays.sort() to sort them

String str1 = "anagram";
String str1 = "margana";
char arrChar1[]= str1.toLowerCase().toCharArray(); 
char arrChar2[]= str2.toLowerCase().toCharArray();
Arrays.sort(arrChar1);
Arrays.sort(arrChar2);
System.out.println(Arrays.equals(arrChar1,arrChar2));//check if is equals with char to char
//if true is anagram if false is not