Leetcode: Valid Anagram

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

For example,
s = "anagram", t = "nagaram", return true.
s = "rat", t = "car", return false.

Note:
You may assume the string contains only lowercase alphabets.

Follow up:
What if the inputs contain unicode characters? How would you adapt your solution to such case?

还不知道Unicode 该怎么做

只有lowercase alphabets的做法无外乎sort, hashmap, array, bitmap之类的

 Better:

public class Solution {
    public boolean isAnagram(String s, String t) {
        int[] alphabet = new int[26];
        for (int i = 0; i < s.length(); i++) alphabet[s.charAt(i) - 'a']++;
        for (int i = 0; i < t.length(); i++) alphabet[t.charAt(i) - 'a']--;
        for (int i : alphabet) if (i != 0) return false;
        return true;
    }
}

 Unicode Follow up

In Java, a Unicode could be represented by a single char(BMP, Basic Multilingual Plane) or two chars (high surrogate). Bascially, we can use

 

• String.codePointAt(int index) method to get the integer representation of a Unicode (as the key in the hash table)
• and use Character.charCount(int code) to count how many the characters are used there (to correctly increase our index)

public class Solution {
    public boolean isAnagram(String s, String t) {
        if (s==null && t==null) return true;
        else if (s==null || t==null) return false;
        else if (s.length() != t.length()) return false;
        
        Map<Integer, Integer> dict = new HashMap<>();
        int index = 0;
        while(index < s.length()) {
            int charCode = s.codePointAt(index); // Get the integer representation of Unicode 
            dict.put(charCode, dict.getOrDefault(charCode, 0) + 1);
            index += Character.charCount(charCode); // The Unicode could be represented by either one char or two chars
        }
        
        index = 0;
        while(index < t.length()) {
            int charCode = t.codePointAt(index);
            int count = dict.getOrDefault(charCode, 0);
            
            if (count == 0) return false;
            else dict.put(charCode, count - 1);
            
            index += Character.charCount(charCode);
        }
        
        return true;
    }
}