[LeetCode] 0033. Search in Rotated Sorted Array

Search in Rotated Sorted Array

Explore diverse LeetCode solutions in Python, C++, JavaScript, SQL, and TypeScript. Ideal for interview prep, learning, and code practice in multiple programming languages. Github Repo Link

內容目錄

Medium

 


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

Python

				
					from typing import List


class Solution:
    def search(self, nums: List[int], target: int) -> int:
        # return nums.index(target) if target in nums else -1
        n = len(nums)
        left, right = 0, n-1

        while left <= right:
            mid = left + (right - left) // 2
            if nums[mid] > nums[-1]:
                left = mid + 1
            else:
                right = mid - 1

        def binarySearch(left_boundary: int, right_boundary: int, target: int):
            left, right = left_boundary, right_boundary
            while left <= right:
                mid = left + (right - left) // 2
                if nums[mid] == target:
                    return mid
                elif nums[mid] > target:
                    right = mid - 1
                else:
                    left = mid + 1
            return -1

        if (answer := binarySearch(0, left-1, target)) != -1:
            return answer
        return binarySearch(left, n-1, target)
				
			
zh_TW繁體中文