refactor 219

fishercoder1534 · fishercoder1534 · commit 99e4cac64d2f · 2018-09-27T07:56:04.000-07:00
diff --git a/src/main/java/com/fishercoder/solutions/_219.java b/src/main/java/com/fishercoder/solutions/_219.java
@@ -4,26 +4,28 @@
 import java.util.Map;
 
 /**219. Contains Duplicate II
- *
-Given an array of integers and an integer k, find out whether there are two distinct indices i and j in the array such that nums[i] = nums[j] and the difference between i and j is at most k.*/
+
+ Given an array of integers and an integer k,
+ find out whether there are two distinct indices i and j in
+ the array such that nums[i] = nums[j] and the difference between i and j is at most k.
+ */
 public class _219 {
 
-    //pretty straightforward, use a hashmap, key is the number itself, value is the last index that this value appeared in the array
-    //we can keep updating the value as we move forward, since if the current index minus the last index cannot be smaller than k, then
-    //the later indices won't even do either. So, we only need to keep one index in the value of the HashMap. Cheers!
-    public boolean containsNearbyDuplicate(int[] nums, int k) {
-        Map<Integer, Integer> map = new HashMap();
-        for (int i = 0; i < nums.length; i++) {
-            if (map.containsKey(nums[i])) {
-                if (i - map.get(nums[i]) <= k) {
-                    return true;
+    public static class Solution1 {
+        public boolean containsNearbyDuplicate(int[] nums, int k) {
+            Map<Integer, Integer> map = new HashMap();
+            for (int i = 0; i < nums.length; i++) {
+                if (map.containsKey(nums[i])) {
+                    if (i - map.get(nums[i]) <= k) {
+                        return true;
+                    } else {
+                        map.put(nums[i], i);
+                    }
                 } else {
                     map.put(nums[i], i);
                 }
-            } else {
-                map.put(nums[i], i);
             }
+            return false;
         }
-        return false;
     }
 }
