LeetCode033. Search in Rotated Sorted Array

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本文介绍了一种变形的二分查找算法，用于在未知旋转点的升序数组中搜索目标值。算法复杂度为O(log n)，适用于无重复元素的数组。通过调整二分查找的判断条件，实现在旋转数组中的高效搜索。

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

(i.e., [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]).

You are given a target value to search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array.

Your algorithm's runtime complexity must be in the order of O(log n).

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

一种变形的二分查找.方法和二分查找一样,不同的是判断条件多了一些

class Solution(object):
    def search(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: int
        """
        left,right = 0,len(nums)-1
        while left<=right:
            mid = (left+right)//2
            if nums[mid]==target:
                return mid
            if nums[left]<=target<nums[mid] or nums[mid]<nums[left]<=target or target<nums[mid]<nums[left]:
                right = mid-1
            else:
                left = mid+1
        return -1