refactor 137

fishercoder1534 · fishercoder1534 · commit 9b4b4e3558d4 · 2018-04-26T08:37:25.000-07:00
diff --git a/src/main/java/com/fishercoder/solutions/_137.java b/src/main/java/com/fishercoder/solutions/_137.java
@@ -3,45 +3,45 @@
 import java.util.HashMap;
 import java.util.Map;
 
-/**137. Single Number II
+/**
+ * 137. Single Number II
+
 Given an array of integers, every element appears three times except for one. Find that single one.
 
 Note:
 Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
-
 */
 public class _137 {
 
-    public static class Solution1 {
-        public int singleNumber(int[] nums) {
-            Map<Integer, Integer> map = new HashMap();
-            for (int i : nums) {
-                map.put(i, map.getOrDefault(i, 0) + 1);
-            }
-            for (int key : map.keySet()) {
-                if (map.get(key) != 3) {
-                    return key;
-                }
-            }
-            return 0;
+  public static class Solution1 {
+    public int singleNumber(int[] nums) {
+      Map<Integer, Integer> map = new HashMap();
+      for (int i : nums) {
+        map.put(i, map.getOrDefault(i, 0) + 1);
+      }
+      for (int key : map.keySet()) {
+        if (map.get(key) != 3) {
+          return key;
         }
+      }
+      return 0;
     }
+  }
 
-    public static class Solution2 {
-        /**Credit: https://discuss.leetcode.com/topic/11877/detailed-explanation-and-generalization-of-the-bitwise-operation-method-for-single-numbers/2*/
-        public int singleNumber(int[] nums) {
-            int counter1 = 0;
-            int counter2 = 0;
-            int mask = 0;
-            for (int num : nums) {
-                counter2 ^= counter1 & num;
-                counter1 ^= num;
-                mask = ~(counter1 & counter2);
-                counter1 &= mask;
-                counter2 &= mask;
-            }
-            return counter1;
-        }
+  public static class Solution2 {
+    /** Credit: https://discuss.leetcode.com/topic/11877/detailed-explanation-and-generalization-of-the-bitwise-operation-method-for-single-numbers/2 */
+    public int singleNumber(int[] nums) {
+      int counter1 = 0;
+      int counter2 = 0;
+      int mask = 0;
+      for (int num : nums) {
+        counter2 ^= counter1 & num;
+        counter1 ^= num;
+        mask = ~(counter1 & counter2);
+        counter1 &= mask;
+        counter2 &= mask;
+      }
+      return counter1;
     }
-
+  }
 }
