在 Valid Anagram 程序中未通过所有测试用例

Question

Given two strings s and t, write a function to determine if t is an anagram of s.

Example 1:

Input: s = "anagram", t = "nagaram" Output: true Example 2:

Input: s = "rat", t = "car" Output: false

example 3 "aad" "cab" Output true Expected false

my 3 test case is giving output true why?

class Solution {
    public boolean isAnagram(String s, String t) {
        if (s.isEmpty() && t.isEmpty()) {
            return true;
        }
        if (s.length() != t.length()) {
            return false;
        }
        char[] a = s.toCharArray();
        char[] b = t.toCharArray();

        Arrays.sort(a);
        Arrays.sort(b);
        for (int i = 0; i <= a.length; i++) {
            for (int j = 0; j <= b.length; j++) {
                if (a[i] == b[j]) {
                    return true;    
                }
            }
        }
        return false;
    }
}

Answer 1

By using a nested for loop, you will iterate over every possible pair (i, j) with i and j an in idex in a and b respectively. Furthermore you use i++ and j++ twice, and thus you will skip the even indices. You can not return true from the moment a[i++] == b[j++] matches. In order to know if something is an anagram, you need to iterate over all elements. You can return false from the moment a[i] != b[i] however. Finally the bound should be i < a.length , not i <= a.length .

You thus need one for loop where you make a single increment and compare a[i] with b[i] :

public boolean isAnagram(String s, String t) {
    if(s.length() != t.length()){
        return false;
    }
    char[] a = s.toCharArray();
    char[] b = t.toCharArray();

    Arrays.sort(a);
    Arrays.sort(b);

    for(int i = 0; i < a.length; i++) {
        if() {
            return ;
        }
    }
    return ;
}

Answer 2

You are just comparing the first letter of cab and rat which is a , and returning True, actually you need to check all letters.

Hence make the condition negative and swap the returns.

 if(a[i++] !=b[j++]){
   return false;
  } 
return true; 

The above code will return false, when characters aren't equal, hence the last line is reached only when all chars are equal.