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

1) 这个题目思路就是先画图, 分区间讨论,注意是左移还是右移,最好画图判断下。

2)可以利用[LeetCode] 153. Find Minimum in Rotated Sorted Array_Medium tag: Binary Search 的思路,通过lg n 找到转折点,然后分别在两段里面找。

T: O(lgn)

Code

1)

class Solution:
def search(self, nums, target):
if not nums: return -1
S, E, l, r = nums[0], nums[-1], 0, len(nums) - 1
while l + 1 < r:
mid = l + (r - l)//2
if target == nums[mid]: return mid
if nums[mid] >= S:
if S <= target <= nums[mid]:
r = mid
else:
l = mid
else:
if nums[mid] <= target <= E:
l = mid
else:
r = mid
if nums[l] == target: return l
if nums[r] == target: return r
return -1

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