33. Search in Rotated Sorted Array
Array, Binary Search ·Problem Statement
link: https://leetcode.com/problems/search-in-rotated-sorted-array/description/ https://leetcode.cn/problems/search-in-rotated-sorted-array/description/
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:
Input: nums = [4,5,6,7,0,1,2]
Output: 0
Input: nums = [4,5,6,7,0,1,2]
Output: 3
Input: nums = [1]
Output: 0
Solution Approach
The key is to utilize binary search while determining the sorted half of the rotated array to narrow down the search for the target.
Algorithm
- Using binary search, identify which half of the rotated array, left or right of the middle, is sorted.
- Determine if the target lies within the sorted half’s range and focus the search there, otherwise switch to the opposite half.
Implement
class Solution:
def search(self, nums: List[int], target: int) -> int:
if not nums:
return -1
l, r = 0, len(nums) - 1
while l <= r:
mid = (l + r)// 2
if nums[mid] == target:
return mid
if nums[0] <= nums[mid]:
if nums[0] <= target < nums[mid]:
r = mid - 1
else:
l = mid + 1
else:
if nums[mid] < target <= nums[len(nums) - 1]:# notice: here is len(nums) - 1
l = mid + 1
else:
r = mid - 1
return -1