There is an integer array nums sorted in ascending order (with distinct values).
Prior to being passed to your function, nums is possibly rotated at an unknown pivot index k (1 <= k < nums.length) such that the resulting array is [nums[k], nums[k+1], …, nums[n-1], nums[0], nums[1], …, nums[k-1]] (0-indexed). For example, [0,1,2,4,5,6,7] might be rotated at pivot index 3 and become [4,5,6,7,0,1,2].
Given the array nums after the possible rotation and an integer target, return the index of target if it is in nums, or -1 if it is not in nums.
You must write an algorithm with O(log n) runtime complexity.
Example 1:
Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4
Example 2:
Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1
Example 3:
Input: nums = [1], target = 0
Output: -1
Constraints:
1 <= nums.length <= 5000
-104 <= nums[i] <= 104
All values of nums are unique.
nums is an ascending array that is possibly rotated.
-104 <= target <= 104
https://leetcode.cn/problems/search-in-rotated-sorted-array/description/
方法一：两次二分
class Solution:# 153. 寻找旋转排序数组中的最小值def findMin(self, nums: List[int]) -> int:left, right = -1, len(nums) - 1 # 开区间 (-1, n-1)while left + 1 < right: # 开区间不为空mid = (left + right) // 2if nums[mid] < nums[-1]: # 蓝色right = midelse: # 红色left = midreturn right# 有序数组中找 target 的下标def lower_bound(self, nums: List[int], left: int, right: int, target: int) -> int:while left + 1 < right: # 开区间不为空mid = (left + right) // 2# 循环不变量：# nums[left] < target# nums[right] >= targetif nums[mid] < target:left = mid # 范围缩小到 (mid, right)else:right = mid # 范围缩小到 (left, mid)return right if nums[right] == target else -1def search(self, nums: List[int], target: int) -> int:i = self.findMin(nums)if target > nums[-1]: # target 在第一段return self.lower_bound(nums, -1, i, target) # 开区间 (-1, i)# target 在第二段return self.lower_bound(nums, i - 1, len(nums), target) # 开区间 (i-1, n)
方法二：一次二分
class Solution:def search(self, nums: List[int], target: int) -> int:def is_blue(i: int) -> bool:x = nums[i]if x > nums[-1]:return target > nums[-1] and x >= targetreturn target > nums[-1] or x >= targetleft, right = -1, len(nums) - 1 # 开区间 (-1, n-1)while left + 1 < right: # 开区间不为空mid = (left + right) // 2if is_blue(mid):right = midelse:left = midreturn right if nums[right] == target else -1