refactor 266

Author: fishercoder1534

src/main/java/com/fishercoder/solutions/_266.java

@@ -3,40 +3,31 @@
 import java.util.HashMap;
 import java.util.Map;
 
-/**Given a string, determine if a permutation of the string could form a palindrome.
-
- For example,
- "code" -> False, "aab" -> True, "carerac" -> True.
-
- Hint:
-
- Consider the palindromes of odd vs even length. What difference do you notice?
- Count the frequency of each character.
- If each character occurs even number of times, then it must be a palindrome. How about character which occurs odd number of times?*/
 public class _266 {
 
-    public boolean canPermutePalindrome(String s) {
-
-        char[] chars = s.toCharArray();
-        Map<Character, Integer> map = new HashMap<Character, Integer>();
-        for (char c : chars) {
-            if (!map.containsKey(c)) {
-                map.put(c, 1);
-            } else {
-                map.put(c, map.get(c) + 1);
+    public static class Solution1 {
+        public boolean canPermutePalindrome(String s) {
+
+            char[] chars = s.toCharArray();
+            Map<Character, Integer> map = new HashMap<>();
+            for (char c : chars) {
+                if (!map.containsKey(c)) {
+                    map.put(c, 1);
+                } else {
+                    map.put(c, map.get(c) + 1);
+                }
             }
-        }
-        int evenCount = 0;
-        for (Map.Entry<Character, Integer> e : map.entrySet()) {
-            if (e.getValue() % 2 != 0) {
-                evenCount++;
-            }
-            if (evenCount > 1) {
-                return false;
+            int evenCount = 0;
+            for (Map.Entry<Character, Integer> e : map.entrySet()) {
+                if (e.getValue() % 2 != 0) {
+                    evenCount++;
+                }
+                if (evenCount > 1) {
+                    return false;
+                }
             }
+            return true;
         }
-        return true;
-
     }
 
 }