242. 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?
Solution if characters are ASCII, this method works
7ms 63.99%

public class Solution {
public boolean isAnagram(String s, String t) {
if(s.length() != t.length()){
return false;
}
int array [] = new int [256];
for(int i = 0; i < s.length(); i++){
int num = s.charAt(i) - 'a';
array[num]++;
}

for(int j = 0; j < t.length(); j++){
int n = t.charAt(j) - 'a';
array
--;
if(array
< 0){
return false;
}
}
return true;
}

}

Solution 2 Sort the string and compare

5ms 88.98%

public class Solution {
public boolean isAnagram(String s, String t) {
if(s.length() != t.length()){
return false;
}
return sort(s).equals(sort(t));
}
public String sort(String s) {
char[] content = s.toCharArray();
Arrays.sort(content);
return new String(content);
}
}

Solution3  Use HashMap

54ms 12.50%

public class Solution {
public boolean isAnagram(String s, String t) {
if(s.length() != t.length()){
return false;
}
Map<Character, Integer> map = new HashMap<Character, Integer>();
for(int i = 0; i < s.length(); i++){
if(!map.containsKey(s.charAt(i))){
map.put(s.charAt(i), 1);
}else{
map.put(s.charAt(i), map.get(s.charAt(i)) + 1);
}
}
for(int i = 0; i < t.length(); i++){
if(!map.containsKey(t.charAt(i))){
return false;
}else{
i
4000
nt c = map.get(t.charAt(i));
c--;
if(c == 0){
map.remove(t.charAt(i));
}else{
map.put(t.charAt(i), c);
}
}
}
return true;
}
}