[leetcode] 33. Search in Rotated Sorted Array

题目

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
• -10^4 <= nums[i] <= 10^4
• All values of nums are unique.
• nums is an ascending array that is possibly rotated.
• -10^4 <= target <= 10^4

思路

两次二分查找，第一次查找数组反转的边界，将数组分隔为左右两个数组。在根据目标值的大小判断是使用左数组还是右数组后，再进行一次二分查找找出目标值的位置。

代码

python版本：

class Solution:
    def search(self, nums: List[int], target: int) -> int:
        l, r = 0, len(nums)
        while l < r:
            mid = (l+r)//2
            if nums[mid] >= nums[0]:
                l = mid+1
            else:
                r = mid
        if target >= nums[0]:
            l = 0
        else:
            r = len(nums)
        while l < r:
            mid = (l+r)//2
            if nums[mid] > target:
                r = mid
            elif nums[mid] < target:
                l = mid+1
            else:
                return mid
        return -1