嗯第一眼就感觉是二分查找法，然后我就一通乱写，就写完了，时间复杂度是O（logN），空间复杂度是O（1）。不过我这个是递归版本的而且好像并没有利用有序这个特性，所以我就又复制了一版用循环实现的。
class Solution:
def search1(self, nums: List[int], target: int) -> int:
self.ans = -1
left, right = 0, len(nums)-1
self.find(nums, target, left, right)
return self.ans
def find(self, nums, target, left, right):
mid = (left + right) // 2
if nums[mid] == target:
self.ans = mid
return
if left >= right:
return
self.find(nums, target, left, mid)
self.find(nums, target, mid+1, right)
def search2(self, nums: List[int], target: int) -> int:
if not nums:
return -1
l, r = 0, len(nums) - 1
while l <= r:
mid = (l + r) // 2
if nums[mid] == target:
return mid
if nums[0] <= nums[mid]:
if nums[0] <= target < nums[mid]:
r = mid - 1
else:
l = mid + 1
else:
if nums[mid] < target <= nums[len(nums) - 1]:
l = mid + 1
else:
r = mid - 1
return -1PREVIOUSmq014.下一个排列