- Search in Rotated Sorted Array Medium
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Share 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
思路：时间复杂度为O(log n)，用二分查找 先判断左边，右边哪边有序 有序的一边是否包含target，包含，l=mid+1 or r=mid-1
代码：python3
class Solution:
def search(self, nums: List[int], target: int) -> int:
if not nums:
return -1
if len(nums)==1:
if target==nums[0]:
return 0
else:
return -1
i=0
j=len(nums)-1
while i<j:
mid = (i+j)//2
print(mid)
if nums[i]==target:
return i
if nums[j]==target:
return j
if nums[mid]==target:
return mid
if nums[i]<nums[mid]:
if nums[i]<target and nums[mid]>target:
j=mid-1
else:
i=mid+1
print(str(i)+"#"+str(j))
else:
if nums[j]>target and nums[mid]<target:
i=mid+1
else:
j=mid-1
print(str(i)+"*"+str(j))
return -1
