[LeetCode]题解（python）：033-Search in Rotated Sorted Array

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题目来源

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https://leetcode.com/problems/search-in-rotated-sorted-array/

Suppose a sorted array 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.

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题意分析

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Input: a list with rotated sorted array with a unknown pivot

Output:index or -1

Conditions:在一个翻转数组中，找到target的位置，若不能找到则返回-1

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题目思路

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此题是关于二分查找的题目， 关键在于确定二分的位置，注意到数组仅是关于一个pivot位置的翻转，意味着这个pivot之前的值都是大于pivot之后的值的，所以在二分的时候利用好这个信息就好了

1. 用first和last存储首尾位置，mid = （first + mid）// 2；
2. 如果target等于nums[mid]，返回target值；
3. 如果nums[first] < nums[mid]:
4. 如果nums[first] >= nums[mid]
5. 对3，4两种条件，加上target与nums[mid]的值的比较，与target与nums[first]的比较，则可确定新的first与last的范围

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AC代码（Python）

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 class Solution(object):
     def search(self, nums, target):
         """
         :type nums: List[int]
         :type target: int
         :rtype: int
         """
         size = len(nums)
         first = 0
         last = size
         while first < last:

             mid = (last + first) // 2
             print(first,mid, last)
             if nums[mid] == target:
                 return mid
             elif nums[first] <= nums[mid]:
                 if target <= nums[mid] and target >= nums[first]:
                     last = mid
                 else:
                     first = mid + 1
             else:
                 if target > nums[mid] and target < nums[first]:
                     first = mid + 1
                 else:
                     last = mid

         return -1