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

LeetCode：連結

變體：LeetCode81：Search in Rotated Sorted Array II

這題和劍指Offer_程式設計題06：旋轉陣列的最小數字(指標)

• 若nums[mid] == target 沒什麼好說的，返回mid
• 若nums[mid] < target
  • 若nums[mid] < nums[right] && target <= nums[right] ||  nums[mid] > nums[right]
    • 即右邊有序且target在右邊的範圍內或者右邊非有序則 left = mid + 1
    • 因為nums[mid] < target 所以若右邊非有序，則左邊一定有序，target一定大於左邊的所有元素，只能在右邊
  • 否則right = mid – 1
• target  < nums[mid]
  • 若nums[left] < nums[mid] && nums[left] <=target
    • 即左邊有序且target在左邊的範圍內  或者 左邊無序列  right = mid – 1;
    • 因為target  < nums[mid] ，若左邊無序則nums[mid]小於右邊的所有元素，只能在左邊。
  • 否則left = mid + 1

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