refactor 409

fishercoder1534 · fishercoder1534 · commit 88fffe63f75d · 2020-07-07T06:43:45.000-07:00
diff --git a/src/main/java/com/fishercoder/solutions/_409.java b/src/main/java/com/fishercoder/solutions/_409.java
@@ -1,48 +1,29 @@
 package com.fishercoder.solutions;
 
-/**
- * 409. Longest Palindrome
- *
- * Given a string which consists of lowercase or uppercase letters, find the length of the longest palindromes that can be built with those letters.
- * This is case sensitive, for example "Aa" is not considered a palindrome here.
- *
- * Note:
- * Assume the length of given string will not exceed 1,010.
- *
- * Example:
- * Input:
- * "abccccdd"
- *
- * Output:
- * 7
- *
- * Explanation:
- * One longest palindrome that can be built is "dccaccd", whose length is 7.
- */
 public class _409 {
-  public static class Solution1 {
-    public int longestPalindrome(String s) {
-      int[] counts = new int[56];
-      for (char c : s.toCharArray()) {
-        if (Character.isUpperCase(c)) {
-          counts[c - 'A' + 33]++;
-        } else {
-          counts[c - 'a']++;
+    public static class Solution1 {
+        public int longestPalindrome(String s) {
+            int[] counts = new int[56];
+            for (char c : s.toCharArray()) {
+                if (Character.isUpperCase(c)) {
+                    counts[c - 'A' + 33]++;
+                } else {
+                    counts[c - 'a']++;
+                }
+            }
+            boolean hasOdd = false;
+            int len = 0;
+            for (int i = 0; i < 56; i++) {
+                if (counts[i] % 2 != 0) {
+                    hasOdd = true;
+                    if (counts[i] > 1) {
+                        len += counts[i] - 1;
+                    }
+                } else {
+                    len += counts[i];
+                }
+            }
+            return hasOdd ? len + 1 : len;
         }
-      }
-      boolean hasOdd = false;
-      int len = 0;
-      for (int i = 0; i < 56; i++) {
-        if (counts[i] % 2 != 0) {
-          hasOdd = true;
-          if (counts[i] > 1) {
-            len += counts[i] - 1;
-          }
-        } else {
-          len += counts[i];
-        }
-      }
-      return hasOdd ? len + 1 : len;
     }
-  }
 }
