refactor 259

Author: fishercoder1534

src/main/java/com/fishercoder/solutions/_259.java

@@ -2,41 +2,32 @@
 
 import java.util.Arrays;
 
-/**259. 3Sum Smaller
-
-Given an array of n integers nums and a target, find the number of index triplets i, j, k with 0 <= i < j < k < n that satisfy the condition nums[i] + nums[j] + nums[k] < target.
-
-For example, given nums = [-2, 0, 1, 3], and target = 2.
-
-Return 2. Because there are two triplets which sums are less than 2:
-
-[-2, 0, 1]
-[-2, 0, 3]
-
-Follow up:
-Could you solve it in O(n2) runtime? */
 public class _259 {
 
-    /**Basically, very similar to 3Sum, but the key is that you'll have to add result by (right-left), not just increment result by 1!*/
-    public int threeSumSmaller(int[] nums, int target) {
-        if (nums == null || nums.length == 0) {
-            return 0;
-        }
-        int result = 0;
-        Arrays.sort(nums);
-        for (int i = 0; i < nums.length - 2; i++) {
-            int left = i + 1;
-            int right = nums.length - 1;
-            while (left < right) {
-                int sum = nums[i] + nums[left] + nums[right];
-                if (sum < target) {
-                    result += right - left;//this line is key!
-                    left++;
-                } else {
-                    right--;
+    public static class Solution1 {
+        /**
+         * Basically, very similar to 3Sum, but the key is that you'll have to add result by (right-left), not just increment result by 1!
+         */
+        public int threeSumSmaller(int[] nums, int target) {
+            if (nums == null || nums.length == 0) {
+                return 0;
+            }
+            int result = 0;
+            Arrays.sort(nums);
+            for (int i = 0; i < nums.length - 2; i++) {
+                int left = i + 1;
+                int right = nums.length - 1;
+                while (left < right) {
+                    int sum = nums[i] + nums[left] + nums[right];
+                    if (sum < target) {
+                        result += right - left;//this line is key!
+                        left++;
+                    } else {
+                        right--;
+                    }
                 }
             }
+            return result;
         }
-        return result;
     }
 }