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).
解题思路
和没有被rotate过的一样，使用二分的方法，我们总考虑二分时候的左(L)右(R)端点，中点(M)和目标(T)
L <= M < T, T < L <= M, M < T < L，当这三种情况出现时，移动L到M+1，其他情况移动R到M-1
代码
class Solution:
def search(self, nums: List[int], target: int) -> int:
l, r = 0, len(nums) - 1
while l <= r:
m = (l + r) // 2
if nums[m] == target:
return m
elif nums[l] <= nums[m] < target or \
target < nums[l] <= nums[m] or \
nums[m] < target < nums[l]:
l = m + 1
else:
r = m - 1
return -1
