refactor 154

fishercoder1534 · fishercoder1534 · commit 5b8c58d589c2 · 2020-07-25T06:31:04.000-07:00
diff --git a/src/main/java/com/fishercoder/solutions/_154.java b/src/main/java/com/fishercoder/solutions/_154.java
@@ -1,52 +1,28 @@
 package com.fishercoder.solutions;
 
-/**
- * 154. Find Minimum in Rotated Sorted Array II
-
- Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
-
- (i.e.,  [0,1,2,4,5,6,7] might become  [4,5,6,7,0,1,2]).
-
- Find the minimum element.
-
- The array may contain duplicates.
-
- Example 1:
- Input: [1,3,5]
- Output: 1
-
- Example 2:
- Input: [2,2,2,0,1]
- Output: 0
-
- Note:
- This is a follow up problem to Find Minimum in Rotated Sorted Array.
- Would allow duplicates affect the run-time complexity? How and why?
-
- */
 public class _154 {
-  public static class Solution1 {
-    public int findMin(int[] nums) {
-      int left = 0;
-      int right = nums.length - 1;
-      if (nums[left] < nums[right]) {
-        return nums[left];
-      }
-      int min = nums[0];
-      while (left + 1 < right) {
-        int mid = left + (right - left) / 2;
-        min = Math.min(min, nums[mid]);
-        if (nums[mid] > nums[left]) {
-          min = Math.min(nums[left], min);
-          left = mid + 1;
-        } else if (nums[mid] < nums[left]) {
-          right = mid - 1;
-        } else {
-          left++;
+    public static class Solution1 {
+        public int findMin(int[] nums) {
+            int left = 0;
+            int right = nums.length - 1;
+            if (nums[left] < nums[right]) {
+                return nums[left];
+            }
+            int min = nums[0];
+            while (left + 1 < right) {
+                int mid = left + (right - left) / 2;
+                min = Math.min(min, nums[mid]);
+                if (nums[mid] > nums[left]) {
+                    min = Math.min(nums[left], min);
+                    left = mid + 1;
+                } else if (nums[mid] < nums[left]) {
+                    right = mid - 1;
+                } else {
+                    left++;
+                }
+            }
+            min = Math.min(min, Math.min(nums[left], nums[right]));
+            return min;
         }
-      }
-      min = Math.min(min, Math.min(nums[left], nums[right]));
-      return min;
     }
-  }
 }
