【LEETCODE】33-Search in Rotated Sorted Array [Python]

Suppose a sorted array 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.

题意：
假设一个排好的数组，在某一个节点处逆转了
(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

给定一个目标值，如果数组里有此值，则返回它的下标，如果没有，则返回－1
可以假设数组里没有重复

思路：
二分法，先判断中值是属于哪边的升序序列，再根据端点值继续判断 target 该落在左边还是右边区域

Python：

class Solution(object):
def search(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: int
"""

start=0
end=len(nums)-1

while start<=end:

mid=(start+end)/2

if nums[mid]==target:
return mid

if nums[mid]>=nums[start]:                              #当nums[mid]属于左边升序序列时

if target>=nums[start] and target<nums[mid]:
end=mid-1
else:
start=mid+1

if nums[mid]<nums[end]:                                 #当nums[mid]属于右边升序序列时

if target>nums[mid] and target<=nums[end]:
start=mid+1
else:
end=mid-1

return -1