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

和剑指offer不一样，那个题是找最小值，而且加难度的地方是这个数组是不递减的

其实自己的思路对，但是逻辑混乱导致WA

想着如果mid在左边，target在l mid中间就是r=m-1 反之l=m+1

如果mid在右边，target在mid r中间就是l=m+1 反之r=m-1

class Solution(object):
    def search(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: int
        """
        l,r=0,len(nums)-1
        while(l<=r):
            m=(l+r)/2
            if(nums[m]==target):
                return m
            if(nums[l]<=nums[m]):
                if(target<=nums[m] and target>=nums[l]):
                    r=m-1
                else:
                    l=m+1
            else:
                if(target>=nums[m] and target<=nums[r]):
                    l=m+1
                else:
                    r=m-1
        return -1