33. Search in Rotated Sorted Array--python

最新推荐文章于 2020-06-28 12:02:28 发布

原创最新推荐文章于 2020-06-28 12:02:28 发布 · 584 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode

leetcode 专栏收录该内容

13 篇文章

订阅专栏

本文介绍了一种在旋转排序数组中寻找特定元素的高效算法。通过二分搜索法，在O(log n)的时间复杂度内找到目标值的位置。文章详细解析了如何判断搜索范围并给出了具体实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目：

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

解题：

二分搜索法的关键在于获得了中间数后，判断下面要搜索左半段还是右半段，如果中间的数小于最右边的数，则右半段是有序的，若中间数大于最右边数，则左半段是有序的，我们只要在有序的半段里用首尾两个数组来判断目标值是否在这一区域内，这样就可以确定保留哪半边了。

class Solution(object):
    def search(self, nums, target):

        l , r = 0 ,len(nums)-1 
        while l <= r :
            mid  = (l+r) //2
            if nums[mid] == target:
                return mid
            if nums[mid] >= nums[l]: 
                if target >= nums[l] and target <= nums[mid]:  
                    r = mid 
                else : 
                    l = mid +1
            if nums[mid] <= nums[r] :
                if target >= nums[mid] and target <= nums[r]:
                    l = mid+1
                else:
                    r= mid 
            
        return -1