edit 242

Author: fishercoder1534

src/main/java/com/fishercoder/solutions/_242.java

@@ -2,7 +2,9 @@
 
 import java.util.Arrays;
 
-/**Given two strings s and t, write a function to determine if t is an anagram of s.
+/**
+ * 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.
@@ -12,28 +14,33 @@
  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?*/
+ What if the inputs contain unicode characters? How would you adapt your solution to such case?
+ */
 
 public class _242 {
-    public boolean isAnagram_solution1(String s, String t) {
-        char[] schar = s.toCharArray();
-        char[] tchar = t.toCharArray();
-        Arrays.sort(schar);
-        Arrays.sort(tchar);
-        return new String(schar).equals(new String(tchar));
-    }
-    
-    //another way: although much slower
-    public boolean isAnagram_solution2(String s, String t) {
-        if(s == null || t == null || s.length() != t.length()) return false;
-        int[] counts = new int[26];
-        for(int i = 0; i < s.length(); i++){
-            counts[s.charAt(i) - 'a']++;
-            counts[t.charAt(i) - 'a']--;
+
+    public static class SortingSolution {
+        public boolean isAnagram(String s, String t) {
+            char[] schar = s.toCharArray();
+            char[] tchar = t.toCharArray();
+            Arrays.sort(schar);
+            Arrays.sort(tchar);
+            return new String(schar).equals(new String(tchar));
         }
-        for(int i : counts){
-            if(i != 0) return false;
+    }
+
+    public static class CountingSolution {
+        public boolean isAnagram(String s, String t) {
+            if (s == null || t == null || s.length() != t.length()) return false;
+            int[] counts = new int[26];
+            for (int i = 0; i < s.length(); i++) {
+                counts[s.charAt(i) - 'a']++;
+                counts[t.charAt(i) - 'a']--;
+            }
+            for (int i : counts) {
+                if (i != 0) return false;
+            }
+            return true;
         }
-        return true;
     }
 }