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