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feat: add solutions to lc/lcci problems (doocs#1660)
* lc No.0081.Search in Rotated Sorted Array II * lcci No.10.03.Search Rotate Array
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‎lcci/10.03.Search Rotate Array/README.md

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## 题目描述
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<!-- 这里写题目描述 -->
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<p>搜索旋转数组。给定一个排序后的数组，包含n个整数，但这个数组已被旋转过很多次了，次数不详。请编写代码找出数组中的某个元素，假设数组元素原先是按升序排列的。若有多个相同元素，返回索引值最小的一个。</p>
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<p><strong>示例1:</strong></p>
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<pre><strong> 输入</strong>: arr = [15, 16, 19, 20, 25, 1, 3, 4, 5, 7, 10, 14], target = 5
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：二分查找**
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我们定义二分查找的左边界 $l=0$，右边界 $r=n-1$，其中 $n$ 为数组的长度。
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每次在二分查找的过程中，我们会得到当前的中点 $mid=(l+r)/2$。
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- 如果 $nums[mid] \gt nums[r]$，说明 $[l,mid]$ 是有序的，此时如果 $nums[l] \le target \le nums[mid]$，说明 $target$ 位于 $[l,mid]$，否则 $target$ 位于 $[mid+1,r]$。
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- 如果 $nums[mid] \lt nums[r]$，说明 $[mid+1,r]$ 是有序的，此时如果 $nums[mid] \lt target \le nums[r]$，说明 $target$ 位于 $[mid+1,r]$，否则 $target$ 位于 $[l,mid]$。
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- 如果 $nums[mid] = nums[r]$，说明元素 $nums[mid]$ 和 $nums[r]$ 相等，此时无法判断 $target$ 位于哪个区间，我们只能将 $r$ 减少 $1$。
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二分查找结束后，如果 $nums[l] = target$，则说明数组中存在目标值 $target$，否则说明不存在。
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注意，如果一开始 $nums[l] = nums[r]$，我们循环将 $r$ 减少 $1$，直到 $nums[l] \ne nums[r]$。
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时间复杂度近似 $O(\log n)$，空间复杂度 $O(1)$。其中 $n$ 为数组的长度。
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相似题目：
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- [81. 搜索旋转排序数组 II](/solution/0000-0099/0081.Search%20in%20Rotated%20Sorted%20Array%20II/README.md)
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def search(self, arr: List[int], target: int) -> int:
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l, r = 0, len(arr) - 1
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while arr[l] == arr[r]:
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r -= 1
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while l < r:
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mid = (l + r) >> 1
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if arr[mid] > arr[r]:
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if arr[l] <= target <= arr[mid]:
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r = mid
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else:
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l = mid + 1
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elif arr[mid] < arr[r]:
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if arr[mid] < target <= arr[r]:
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l = mid + 1
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else:
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r = mid
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else:
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r -= 1
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return l if arr[l] == target else -1
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int search(int[] arr, int target) {
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int l = 0, r = arr.length - 1;
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while (arr[l] == arr[r]) {
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--r;
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}
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while (l < r) {
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int mid = (l + r) >> 1;
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if (arr[mid] > arr[r]) {
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if (arr[l] <= target && target <= arr[mid]) {
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r = mid;
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} else {
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l = mid + 1;
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}
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} else if (arr[mid] < arr[r]) {
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if (arr[mid] < target && target <= arr[r]) {
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l = mid + 1;
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} else {
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r = mid;
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}
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} else {
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--r;
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}
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}
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return arr[l] == target ? l : -1;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int search(vector<int>& arr, int target) {
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int l = 0, r = arr.size() - 1;
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while (arr[l] == arr[r]) {
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--r;
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}
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while (l < r) {
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int mid = (l + r) >> 1;
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if (arr[mid] > arr[r]) {
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if (arr[l] <= target && target <= arr[mid]) {
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r = mid;
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} else {
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l = mid + 1;
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}
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} else if (arr[mid] < arr[r]) {
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if (arr[mid] < target && target <= arr[r]) {
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l = mid + 1;
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} else {
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r = mid;
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}
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} else {
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--r;
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}
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}
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return arr[l] == target ? l : -1;
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}
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};
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```
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### **Go**
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```go
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func search(arr []int, target int) int {
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l, r := 0, len(arr)-1
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for arr[l] == arr[r] {
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r--
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}
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for l < r {
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mid := (l + r) >> 1
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if arr[mid] > arr[r] {
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if arr[l] <= target && target <= arr[mid] {
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r = mid
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} else {
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l = mid + 1
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}
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} else if arr[mid] < arr[r] {
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if arr[mid] < target && target <= arr[r] {
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l = mid + 1
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} else {
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r = mid
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}
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} else {
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r--
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}
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}
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if arr[l] == target {
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return l
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}
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return -1
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}
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```
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### **TypeScript**
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```ts
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function search(arr: number[], target: number): number {
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let [l, r] = [0, arr.length - 1];
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while (arr[l] === arr[r]) {
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--r;
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}
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while (l < r) {
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const mid = (l + r) >> 1;
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if (arr[mid] > arr[r]) {
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if (arr[l] <= target && target <= arr[mid]) {
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r = mid;
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} else {
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l = mid + 1;
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}
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} else if (arr[mid] < arr[r]) {
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if (arr[mid] < target && target <= arr[r]) {
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l = mid + 1;
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} else {
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r = mid;
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}
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} else {
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--r;
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}
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}
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return arr[l] === target ? l : -1;
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}
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```
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### **...**

‎lcci/10.03.Search Rotate Array/README_EN.md

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### **Python3**
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```python
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class Solution:
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def search(self, arr: List[int], target: int) -> int:
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l, r = 0, len(arr) - 1
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while arr[l] == arr[r]:
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r -= 1
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while l < r:
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mid = (l + r) >> 1
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if arr[mid] > arr[r]:
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if arr[l] <= target <= arr[mid]:
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r = mid
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else:
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l = mid + 1
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elif arr[mid] < arr[r]:
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if arr[mid] < target <= arr[r]:
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l = mid + 1
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else:
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r = mid
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else:
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r -= 1
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return l if arr[l] == target else -1
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```
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### **Java**
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```java
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class Solution {
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public int search(int[] arr, int target) {
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int l = 0, r = arr.length - 1;
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while (arr[l] == arr[r]) {
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--r;
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}
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while (l < r) {
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int mid = (l + r) >> 1;
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if (arr[mid] > arr[r]) {
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if (arr[l] <= target && target <= arr[mid]) {
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r = mid;
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} else {
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l = mid + 1;
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}
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} else if (arr[mid] < arr[r]) {
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if (arr[mid] < target && target <= arr[r]) {
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l = mid + 1;
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} else {
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r = mid;
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}
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} else {
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--r;
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}
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}
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return arr[l] == target ? l : -1;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int search(vector<int>& arr, int target) {
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int l = 0, r = arr.size() - 1;
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while (arr[l] == arr[r]) {
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--r;
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}
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while (l < r) {
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int mid = (l + r) >> 1;
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if (arr[mid] > arr[r]) {
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if (arr[l] <= target && target <= arr[mid]) {
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r = mid;
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} else {
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l = mid + 1;
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}
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} else if (arr[mid] < arr[r]) {
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if (arr[mid] < target && target <= arr[r]) {
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l = mid + 1;
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} else {
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r = mid;
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}
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} else {
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--r;
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}
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}
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return arr[l] == target ? l : -1;
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}
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};
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```
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### **Go**
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```go
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func search(arr []int, target int) int {
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l, r := 0, len(arr)-1
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for arr[l] == arr[r] {
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r--
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}
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for l < r {
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mid := (l + r) >> 1
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if arr[mid] > arr[r] {
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if arr[l] <= target && target <= arr[mid] {
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r = mid
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} else {
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l = mid + 1
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}
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} else if arr[mid] < arr[r] {
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if arr[mid] < target && target <= arr[r] {
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l = mid + 1
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} else {
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r = mid
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}
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} else {
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r--
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}
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}
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if arr[l] == target {
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return l
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}
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return -1
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}
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```
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### **TypeScript**
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```ts
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function search(arr: number[], target: number): number {
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let [l, r] = [0, arr.length - 1];
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while (arr[l] === arr[r]) {
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--r;
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}
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while (l < r) {
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const mid = (l + r) >> 1;
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if (arr[mid] > arr[r]) {
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if (arr[l] <= target && target <= arr[mid]) {
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r = mid;
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} else {
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l = mid + 1;
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}
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} else if (arr[mid] < arr[r]) {
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if (arr[mid] < target && target <= arr[r]) {
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l = mid + 1;
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} else {
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r = mid;
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}
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} else {
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--r;
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}
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}
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return arr[l] === target ? l : -1;
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}
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```
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### **...**

‎lcci/10.03.Search Rotate Array/Solution.cpp

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class Solution {
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public:
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int search(vector<int>& arr, int target) {
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int l = 0, r = arr.size() - 1;
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while (arr[l] == arr[r]) {
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--r;
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}
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while (l < r) {
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int mid = (l + r) >> 1;
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if (arr[mid] > arr[r]) {
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if (arr[l] <= target && target <= arr[mid]) {
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r = mid;
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} else {
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l = mid + 1;
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}
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} else if (arr[mid] < arr[r]) {
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if (arr[mid] < target && target <= arr[r]) {
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l = mid + 1;
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} else {
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r = mid;
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}
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} else {
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--r;
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}
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}
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return arr[l] == target ? l : -1;
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}
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};

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