LeetCode 33. Search in Rotated Sorted Array

题目描述

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.

解题思路

看到 O(log n) 的时间复杂度，第一时间想到二分查找。但是给出的列表是局部有序的，我们就需要改变传统的二分的思路。
直接看代码。

代码 (Python 3)

class Solution:
    def search(self, nums, target: int) -> int:
        if target not in nums:
            return -1
        left = 0
        right = len(nums) - 1
        while left <= right:  # 开始二分查找
            mid = (left + right) // 2
            if nums[mid] == target:  # 找到即退出
                return mid
            elif nums[mid] >= nums[left]: # 关键！若此条件满足，则在 [left, mid] 区间为单调
                if nums[left] <= target <= nums[mid]:
                    right = mid
                else:
                    left = mid + 1
            else: 
                if nums[mid] < target <= nums[right]:
                    left = mid + 1
                else:
                    right = mid - 1
        return -1

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来源： https://blog.csdn.net/weixin_43762534/article/details/122042172