動機

用bsearch去猜數字

Problem

Let's call an array arr a mountain if the following properties hold:

  • arr.length >= 3
  • There exists some i with 0 < i < arr.length - 1 such that:
    • arr[0] < arr[1] < ... arr[i-1] < arr[i]
    • arr[i] > arr[i+1] > ... > arr[arr.length - 1]

Given an integer array arr that is guaranteed to be a mountain, return any i such that arr[0] < arr[1] < ... arr[i - 1] < arr[i] > arr[i + 1] > ... > arr[arr.length - 1].

 

Example 1:

Input: arr = [0,1,0]Output: 1

Example 2:

Input: arr = [0,2,1,0]Output: 1

Example 3:

Input: arr = [0,10,5,2]Output: 1

Example 4:

Input: arr = [3,4,5,1]Output: 2

Example 5:

Input: arr = [24,69,100,99,79,78,67,36,26,19]Output: 2

 

Constraints:

  • 3 <= arr.length <= 104
  • 0 <= arr[i] <= 106
  • arr is guaranteed to be a mountain array.

 

Follow up: Finding the O(n) is straightforward, could you find an O(log(n)) solution?

Sol

class Solution:
    def peakIndexInMountainArray(self, arr: List[int]) -> int:
        a,b = 0, len(arr)
        while a<b:
            mid = (a+b)//2
            if arr[mid-1] < arr[mid] > arr[mid+1]:
                return mid
            elif arr[mid-1] < arr[mid] < arr[mid+1]:
                a = mid+1
            elif arr[mid-1] > arr[mid] > arr[mid+1]:
                b = mid
            else:
                raise 'this case wont exist'
        return mid