Problem

There is an integer array nums sorted in ascending order (with distinct values).

Prior to being passed to your function, nums is possibly rotated at an unknown pivot index k (1 <= k < nums.length) such that the resulting array is [nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]] (0-indexed). For example, [0,1,2,4,5,6,7] might be rotated at pivot index 3 and become [4,5,6,7,0,1,2].

Given the array nums after the possible rotation and an integer target, return the index of target if it is in nums, or -1 if it is not in nums.

You must write an algorithm with O(log n) runtime complexity.

Example 1:

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Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4

Example 2:

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Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1

Example 3:

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Input: nums = [1], target = 0
Output: -1

Constraints:

  • 1 <= nums.length <= 5000
  • 10^4 <= nums[i] <= 10^4
  • All values of nums are unique.
  • nums is an ascending array that is possibly rotated.
  • 10^4 <= target <= 10^4

Solve

寫一個 最複雜的，而且還沒用到O(log(n))

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class Solution:
    def search(self, nums: List[int], target: int) -> int:
        if target not in nums:
            return -1

        idx = nums.index(target)

        left = nums[idx:]
        right = nums[:idx]

        b1 = False
        b2 = False

        for i in range( 1 , len(left) ):
            if left[i-1] > left[i]:
                b1 = -1

        for i in range( 1 , len(right) ):
            if right[i-1] > right[i]:
                b2 = -1
                
        if b1 ==-1 and b2 == -1:
            return -1

        return nums.index(target)

Solution2 - Binary Search

改正為 Binary Search

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class Solution:
    def search(self, nums: List[int], target: int) -> int:
        if target not in nums:
            return -1

        left = 0
        right = len(nums)-1

        while left <= right:
            mid = (left+right)//2

            if nums[mid] == target:
                return mid

            if nums[mid] >= nums[left]:
                if nums[left] <= target <= nums[mid]:
                    right = mid - 1
                else:
                    left = mid + 1
            else:
                if nums[mid] <= target <= nums[right]:
                    left = mid + 1
                else:
                    right = mid - 1

        return -1