#33 Search in Rotated Sorted Array——Top 100 Liked Questions-蒲公英云

#33 Search in Rotated Sorted Array——Top 100 Liked Questions

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

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第一次：先找到最大值max(nums)及索引，并将其设为mid，left为0，right为len(nums)-1，这样讲nums分成了两部分，然后通过判断target与nums[left]与nums[right]的大小，判断target在左半部分还是右半部分，然后再对左右两部分应用二叉搜索；复杂度我不会计算，应该是不满足O（log n）的

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class Solution(object):
def search(self, nums, target):
“””
:type nums: List[int]
:type target: int
int
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if len(nums) == 0: return -1
left = 0
right = len(nums) - 1
mid = nums.index(max(nums))
if target == nums[mid]:
return mid
elif target >= nums[left]:
left = 0
right = mid - 1
while left <= right:
mid = (left + right) // 2
if nums[mid] == target:
return mid
elif nums[mid] > target:
right = mid - 1
else:
left = mid + 1
elif target <= nums[right]:
left = mid + 1
right = len(nums) - 1
while left <= right:
mid = (left + right) // 2
if nums[mid] == target:
return mid
elif nums[mid] > target:
right = mid - 1
else:
left = mid + 1
return -1

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Runtime: 24 ms, faster than 66.13% of Python online submissions for Search in Rotated Sorted Array.

Memory Usage: 12.1 MB, less than 5.44% of Python online submissions for Search in Rotated Sorted Array.

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