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
with0 < 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