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[yzx]1006:二分查找_python
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‎TODO.md‎

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## v1
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- [ ] 完善文档细节
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- [ ] python和C++实现
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- [ ] 工程实现用到的算法解析
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- [ ] 周赛计划
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- [ ] 面试体系计划

‎basic_algorithm/binary_search.md‎

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1919

2020
> 给定一个 n 个元素有序的(升序)整型数组 nums 和一个目标值 target ,写一个函数搜索 nums 中的 target,如果目标值存在返回下标,否则返回 -1。
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**C++版本**
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```cpp
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// 二分搜索最常用模板
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int search(vector<int>& nums, int target) {
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return -1;
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}
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```
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**python3版本**
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```python
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class Solution:
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def search(self, nums: List[int], target: int) -> int:
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left = 0
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right = len(nums) - 1
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while left + 1 < right:
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mid = left + (right - left) // 2
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if nums[mid] == target: return mid
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elif nums[mid] < target: left = mid
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else: right = mid
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if nums[left] == target: return left
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if nums[right] == target: return right
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return -1
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```
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大部分二分查找类的题目都可以用这个模板,然后做一点特殊逻辑即可
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```
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## 常见题目
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### [x的平方根](https://leetcode.cn/problems/sqrtx/)
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> 给你一个非负整数 x ,计算并返回 x 的 算术平方根 。
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> 由于返回类型是整数,结果只保留 整数部分 ,小数部分将被 舍去 。
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> 注意:不允许使用任何内置指数函数和算符,例如 pow(x, 0.5) 或者 x ** 0.5 。
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**Python版本**
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```python
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class Solution:
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def mySqrt(self, x: int) -> int:
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if x == 0: return 0
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left, right = 1, x//2 + 1
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while left + 1 < right:
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mid = left + (right - left) // 2
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if mid * mid == x: return mid
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elif mid * mid < x: left = mid
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else: right = mid
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if left * left > x: return left - 1
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elif right * right <= x: return right
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else: return left
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```
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### [猜数字大小](https://leetcode.cn/problems/guess-number-higher-or-lower/)
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> 猜数字游戏的规则如下:
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> 每轮游戏,我都会从 1 到 n 随机选择一个数字。 请你猜选出的是哪个数字。
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> 如果你猜错了,我会告诉你,你猜测的数字比我选出的数字是大了还是小了。
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> 你可以通过调用一个预先定义好的接口 int guess(int num) 来获取猜测结果,返回值一共有 3 种可能的情况(-1,1 或 0):
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> -1:我选出的数字比你猜的数字小 pick < num
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> 1:我选出的数字比你猜的数字大 pick > num
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> 0:我选出的数字和你猜的数字一样。恭喜!你猜对了!pick == num
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**Python版本**
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```python
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# The guess API is already defined for you.
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# @param num, your guess
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# @return -1 if my number is lower, 1 if my number is higher, otherwise return 0
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# def guess(num: int) -> int:
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class Solution:
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def guessNumber(self, n: int) -> int:
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left, right = 1, n
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while left + 1 < right:
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mid = left + (right - left) // 2
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if guess(mid) == 0: return mid
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elif guess(mid) == -1: right = mid
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else: left = mid
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if guess(left) == 0: return left
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else: return right
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```
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### [search-for-range](https://leetcode-cn.com/problems/find-first-and-last-position-of-element-in-sorted-array/)
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思路:核心点就是找第一个 target 的索引,和最后一个 target 的索引,所以用两次二分搜索分别找第一次和最后一次的位置
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**Python3版本**
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```python
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class Solution:
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def searchRange(self, nums: List[int], target: int) -> List[int]:
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if len(nums) == 0: return [-1, -1]
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left, right = 0, len(nums) - 1
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while left + 1 < right:
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mid = left + (right - left) // 2
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if nums[mid] < target: left = mid
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elif nums[mid] > target: right = mid
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else:
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left = mid
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right = mid
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while left > 0 and nums[left-1] == target: left = left - 1
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while right < len(nums) - 1 and nums[right+1] == target: right = right + 1
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return [left, right]
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if nums[left] == target:
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if nums[right] == target: return [left, right]
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else: return [left, left]
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else:
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if nums[right] == target: return [right, right]
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else: return [-1, -1]
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```
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```python
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class Solution:
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def searchRange(self, nums: List[int], target: int) -> List[int]:
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if len(nums) == 0: return [-1, -1]
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'''
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分解成找两次,找最左边一次,找最右边一次
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'''
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left = findTarget(nums, target, True)
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if left == -1: return [-1, -1]
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right = findTarget(nums, target, False)
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return [left, right]
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def findTarget(nums, target, find_left):
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left, right = 0, len(nums) - 1
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while left + 1 < right:
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mid = left + (right - left) // 2
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if nums[mid] < target: left = mid
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elif nums[mid] > target: right = mid
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else:
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if find_left: right = mid
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else: left = mid
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if find_left:
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if nums[left] == target: return left
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elif nums[right] == target: return right
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else: return -1
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else:
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if nums[right] == target: return right
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elif nums[left] == target: return left
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else: return -1
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```
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**C++版本**
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```cpp
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vector<int> searchRange(vector<int>& nums, int target) {
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if(nums.empty()) return {-1, -1};
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> 假设你有 n 个版本 [1, 2, ..., n],你想找出导致之后所有版本出错的第一个错误的版本。
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> 你可以通过调用 bool isBadVersion(version) 接口来判断版本号 version 是否在单元测试中出错。实现一个函数来查找第一个错误的版本。你应该尽量减少对调用 API 的次数。
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**Python3版本**
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```python
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# The isBadVersion API is already defined for you.
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# def isBadVersion(version: int) -> bool:
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class Solution:
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def firstBadVersion(self, n: int) -> int:
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left, right = 1, n
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while left + 1 < right:
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mid = left + (right - left) // 2
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if not isBadVersion(mid): left = mid
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else: right = mid
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if isBadVersion(left): return left
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else: return right
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```
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**C++版本**
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```cpp
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int firstBadVersion(int n) {
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int left = 1, right = n;
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> 假设按照升序排序的数组在预先未知的某个点上进行了旋转( 例如,数组 [0,1,2,4,5,6,7] 可能变为 [4,5,6,7,0,1,2] )。
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> 请找出其中最小的元素。
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**Python版本**
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```python
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class Solution:
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def findMin(self, nums: List[int]) -> int:
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left, right = 0, len(nums) - 1
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if nums[left] <= nums[right]: return nums[left] # 说明还是升序排列
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while left + 1 < right:
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mid = left + (right - left) // 2
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if nums[mid] >= nums[right]: left = mid
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else: right = mid
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return min(nums[left], nums[right])
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```
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**C++版本**
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```cpp
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int findMin(vector<int>& nums) {
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if(nums.empty()) return 0;
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> 搜索一个给定的目标值,如果数组中存在这个目标值,则返回它的索引,否则返回 -1 。
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> 你可以假设数组中不存在重复的元素。
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**Python版本**
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```python
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class Solution:
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def search(self, nums: List[int], target: int) -> int:
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left, right = 0, len(nums) - 1
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while left + 1 < right:
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mid = left + (right - left) // 2
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if nums[mid] == target: return mid
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if nums[mid] > nums[left]:
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if target >= nums[left] and target <= nums[mid]: right = mid
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else: left = mid
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else:
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if target <= nums[right] and target >= nums[mid]: left = mid
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else: right = mid
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if nums[left] == target: return left
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elif nums[right] == target: return right
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else: return -1
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```
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**C++版本**
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```cpp
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int search(vector<int>& nums, int target) {
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if(nums.empty()) return -1;
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> 面试时,可以直接画图进行辅助说明,空讲很容易让大家都比较蒙圈
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### [寻找峰值](https://leetcode.cn/problems/find-peak-element/)
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> 峰值元素是指其值严格大于左右相邻值的元素。
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> 给你一个整数数组 nums,找到峰值元素并返回其索引。数组可能包含多个峰值,在这种情况下,返回 任何一个峰值 所在位置即可。
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> 你可以假设 nums[-1] = nums[n] = -∞ 。
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> 你必须实现时间复杂度为 O(log n) 的算法来解决此问题。
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思路:规定了算法复杂度,那就是二分匹配法
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**Python3版本**
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```python
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class Solution:
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def findPeakElement(self, nums: List[int]) -> int:
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left, right = 0, len(nums) - 1
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while left + 1 < right:
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mid = left + (right - left) // 2
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if nums[mid] < nums[mid+1]: left = mid
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else: right = mid
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if nums[left] < nums[right]: return right
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else: return left
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```
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**前面几道题说明了,二分匹配中,不一定直接对比mid和target, left和mid, mid本身, mid和相邻(mid+1),都是能做compare的。**
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### [search-in-rotated-sorted-array-ii](https://leetcode-cn.com/problems/search-in-rotated-sorted-array-ii/)
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> 假设按照升序排序的数组在预先未知的某个点上进行了旋转。

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