refactor 136

fishercoder1534 · fishercoder1534 · commit e3be701f5cc1 · 2018-04-25T14:48:53.000-07:00
diff --git a/src/main/java/com/fishercoder/solutions/_136.java b/src/main/java/com/fishercoder/solutions/_136.java
@@ -5,6 +5,7 @@
 import java.util.Set;
 
 /**136. Single Number
+
 Given an array of integers, every element appears twice except for one. Find that single one.
 
 Note:
@@ -13,34 +14,37 @@
 
 public class _136 {
 
-    public static class Solution1 {
-        /**approach 1: use set, since this problem explicitly states that every element appears twice and only one appears once
-        so, we could safely remove the ones that are already in the set, O(n) time and O(n) space.
-        HashTable approach works similarly like this one, but it could be more easily extend to follow-up questions.*/
-        public int singleNumber(int[] nums) {
-            Set<Integer> set = new HashSet();
-            for (int i : nums) {
-                if (!set.add(i)) {
-                    set.remove(i);
-                }
-            }
-            Iterator<Integer> it = set.iterator();
-            return it.next();
+  public static class Solution1 {
+    /**
+     * Approach 1: use set, since this problem explicitly states that every element appears twice
+     * and only one appears once so, we could safely remove the ones that are already in the set,
+     * O(n) time and O(n) space. HashTable approach works similarly like this one, but it could be
+     * more easily extend to follow-up questions.
+     */
+    public int singleNumber(int[] nums) {
+      Set<Integer> set = new HashSet();
+      for (int i : nums) {
+        if (!set.add(i)) {
+          set.remove(i);
         }
+      }
+      Iterator<Integer> it = set.iterator();
+      return it.next();
     }
-
-
-    public static class Solution2 {
-        /**approach 2: bit manipulation, use exclusive or ^ to solve this problem:
-        we're using the trick here: every number ^ itself will become zero, so, the only remaining element will be the one that
-        appeared only once.*/
-        public int singleNumber(int[] nums) {
-            int res = 0;
-            for (int i : nums) {
-                res ^= i;
-            }
-            return res;
-        }
+  }
+
+  public static class Solution2 {
+    /**
+     * Approach 2: bit manipulation, use exclusive or ^ to solve this problem: we're using the trick
+     * here: every number ^ itself will become zero, so, the only remaining element will be the one
+     * that appeared only once.
+     */
+    public int singleNumber(int[] nums) {
+      int res = 0;
+      for (int i : nums) {
+        res ^= i;
+      }
+      return res;
     }
-
+  }
 }
