feat: add python and java solution to lcof problem

添加《剑指Offer》题解：面试题11. 旋转数组的最小数字

Author: yanglbme

lcof/README.md

@@ -34,7 +34,11 @@
 │   ├── README.md
 │   ├── Solution.java
 │   └── Solution.py
-└── 面试题10- II. 青蛙跳台阶问题
+├── 面试题10- II. 青蛙跳台阶问题
+│   ├── README.md
+│   ├── Solution.java
+│   └── Solution.py
+└── 面试题11. 旋转数组的最小数字
     ├── README.md
     ├── Solution.java
     └── Solution.py

lcof/面试题11. 旋转数组的最小数字/README.md

@@ -0,0 +1,78 @@
+# [面试题11. 旋转数组的最小数字](https://leetcode-cn.com/problems/xuan-zhuan-shu-zu-de-zui-xiao-shu-zi-lcof/)
+
+## 题目描述
+把一个数组最开始的若干个元素搬到数组的末尾，我们称之为数组的旋转。输入一个递增排序的数组的一个旋转，输出旋转数组的最小元素。例如，数组 `[3,4,5,1,2]` 为 `[1,2,3,4,5]` 的一个旋转，该数组的最小值为 1。  
+
+**示例 1：**
+
+```
+输入：[3,4,5,1,2]
+输出：1
+```
+
+**示例 2：**
+
+```
+输入：[2,2,2,0,1]
+输出：0
+```
+
+## 解法
+### Python3
+```python
+class Solution:
+    def minArray(self, numbers: List[int]) -> int:
+        if len(numbers) == 1 or numbers[0] < numbers[-1]:
+            return numbers[0]
+        left, right = 0, len(numbers) - 1
+        while (right - left > 1):
+            mid = left + ((right - left) >> 1)
+            if numbers[mid] == numbers[left] == numbers[right]:
+                return min(numbers[left: right + 1])
+            if numbers[mid] >= numbers[left]:
+                left = mid
+            elif numbers[mid] <= numbers[right]:
+                right = mid
+        return numbers[right]
+```
+
+### Java
+```java
+class Solution {
+    public int minArray(int[] numbers) {
+        int len = numbers.length;
+        if (len == 1 || numbers[0] < numbers[len - 1]) {
+            return numbers[0];
+        }
+
+        int left = 0, right = len - 1;
+        while (right - left > 1) {
+            int mid = left + ((right - left) >> 1);
+            if (numbers[left] == numbers[mid] && numbers[mid] == numbers[right]) {
+                return findMin(numbers, left, right);
+            }
+            if (numbers[mid] >= numbers[left]) {
+                left = mid;
+            } else if (numbers[mid] <= numbers[right]) {
+                right = mid;
+            }
+        }
+        return numbers[right];
+    }
+
+    private int findMin(int[] numbers, int left, int right) {
+        int min = numbers[left];
+        for (int i = left + 1; i < right; ++i) {
+            if (min > numbers[i]) {
+                min = numbers[i];
+            }
+        }
+        return min;
+    }
+}
+```
+
+### ...
+```
+
+```

lcof/面试题11. 旋转数组的最小数字/Solution.java

@@ -0,0 +1,32 @@
+class Solution {
+    public int minArray(int[] numbers) {
+        int len = numbers.length;
+        if (len == 1 || numbers[0] < numbers[len - 1]) {
+            return numbers[0];
+        }
+
+        int left = 0, right = len - 1;
+        while (right - left > 1) {
+            int mid = left + ((right - left) >> 1);
+            if (numbers[left] == numbers[mid] && numbers[mid] == numbers[right]) {
+                return findMin(numbers, left, right);
+            }
+            if (numbers[mid] >= numbers[left]) {
+                left = mid;
+            } else if (numbers[mid] <= numbers[right]) {
+                right = mid;
+            }
+        }
+        return numbers[right];
+    }
+
+    private int findMin(int[] numbers, int left, int right) {
+        int min = numbers[left];
+        for (int i = left + 1; i < right; ++i) {
+            if (min > numbers[i]) {
+                min = numbers[i];
+            }
+        }
+        return min;
+    }
+}

lcof/面试题11. 旋转数组的最小数字/Solution.py

@@ -0,0 +1,14 @@
+class Solution:
+    def minArray(self, numbers: List[int]) -> int:
+        if len(numbers) == 1 or numbers[0] < numbers[-1]:
+            return numbers[0]
+        left, right = 0, len(numbers) - 1
+        while (right - left > 1):
+            mid = left + ((right - left) >> 1)
+            if numbers[mid] == numbers[left] == numbers[right]:
+                return min(numbers[left: right + 1])
+            if numbers[mid] >= numbers[left]:
+                left = mid
+            elif numbers[mid] <= numbers[right]:
+                right = mid
+        return numbers[right]