refactor 209

fishercoder1534 · fishercoder1534 · commit 78de1497b1a3 · 2018-09-19T07:02:12.000-07:00
diff --git a/src/main/java/com/fishercoder/solutions/_209.java b/src/main/java/com/fishercoder/solutions/_209.java
@@ -1,35 +1,41 @@
 package com.fishercoder.solutions;
 
 /**
- * Given an array of n positive integers and a positive integer s, find the minimal length of a contiguous subarray of which the sum ≥ s. If there isn't one, return 0 instead.
+ * 209. Minimum Size Subarray Sum
+ *
+ * Given an array of n positive integers and a positive integer s,
+ * find the minimal length of a contiguous subarray of which the sum ≥ s.
+ * If there isn't one, return 0 instead.
 
  For example, given the array [2,3,1,2,4,3] and s = 7,
  the subarray [4,3] has the minimal length under the problem constraint.
 
  click to show more practice.
 
  More practice:
- If you have figured out the O(n) solution, try coding another solution of which the time complexity is O(n log n).
+ If you have figured out the O(n) solution,
+ try coding another solution of which the time complexity is O(n log n).
  */
 public class _209 {
 
-    public int minSubArrayLen(int s, int[] nums) {
-        if (nums == null || nums.length == 0) {
-            return 0;
-        }
-        int i = 0;
-        int j = 0;
-        int min = Integer.MAX_VALUE;
-        int sum = 0;
-        while (j < nums.length) {
-            sum += nums[j++];
+    public static class Solution1 {
+        public int minSubArrayLen(int s, int[] nums) {
+            if (nums == null || nums.length == 0) {
+                return 0;
+            }
+            int i = 0;
+            int j = 0;
+            int min = Integer.MAX_VALUE;
+            int sum = 0;
+            while (j < nums.length) {
+                sum += nums[j++];
 
-            while (sum >= s) {
-                min = Math.min(min, j - i);
-                sum -= nums[i++];
+                while (sum >= s) {
+                    min = Math.min(min, j - i);
+                    sum -= nums[i++];
+                }
             }
+            return min == Integer.MAX_VALUE ? 0 : min;
         }
-        return min == Integer.MAX_VALUE ? 0 : min;
     }
-
 }
