Problem 154

Author: YuqiZ2020

0-Binary-Search/154-Find-Minimum-in-Rotated-Sorted-Array-II.java

@@ -0,0 +1,38 @@
+/**
+ * Problem 154: Find Minimum in Rotated Sorted Array II
+ * Prompt: 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.
+ * Date: 07/25/2020
+ */
+class Solution {
+    public int findMin(int[] nums) {
+        int lo = 0;
+        int hi = nums.length - 1;
+        while (lo < hi)
+        {
+            int mid = lo + (hi - lo) / 2;
+            if (nums[mid] < nums[hi])
+                hi = mid;
+            else if (nums[mid] > nums[hi])
+                lo = mid + 1;
+            else
+                hi--;
+        }
+        return nums[lo];
+    }
+}
+
+/**
+ * Notes: There are two situations to consider, one is with rotation, 
+ * and the other is without rotation. We use binary search to look
+ * at which interval we need to look for the smallest number. If
+ * all is with rotation, then we can just divide into two portions, 
+ * if mid < hi then use left, otherwise use right. However, once
+ * we add the without rotation condition. Once we have mid == hi, 
+ * we run into a contradiction of how to deal with it between this
+ * and the previous situation. So we can only do hi-- because we 
+ * know that this current one can be omitted (there are other same
+ * ones in front of it. )
+ */

0-Binary-Search/Binary-Search.md

@@ -49,6 +49,18 @@ class Solution {
 
 ---
 
+### 154. Find Minimum in Rotated Sorted Array II
+
+#### [解法一](154-Find-Minimum-in-Rotated-Sorted-Array-II.java)：二分查找分三类
+
+一共有两种情况，一种是中间某点有置换位置，另一种是完全没有置换位置。如果没有置换位置的话可以分两种情况，如果中间点比右边小则说明可以往左找（包括当前点），否则往右边找（不包括当前点，因为当前点比右边的大或者相等）。
+
+但是如果中间有置换位置，则发现在两数相等的情况下需要往左找，和上一种情况中的往右找相矛盾。所以只能确定的是可以把右侧边界前移（因为前面有和这个数相同的数，所以这个数可以省去）。
+
+![图解](https://raw.githubusercontent.com/YuqiZ2020/PicBed/master/img/20200725203018.png)
+
+---
+
 ### 275. H-Index-II
 #### [解法一](275-H-Index-II.java)：二分查找
 _时间复杂度：O(log(n))； 空间复杂度：O(1)_