A peak element is an element that is greater than its neighbors.
Given an input array nums, where nums[i] ≠ nums[i+1], find a peak element and return its index.
The array may contain multiple peaks, in that case return the index to any one of the peaks is fine.
You may imagine that nums[-1] = nums[n] = -∞.
Define peak A[P] > A[P-1] && A[P] > A[P+1]
deffindPeakElement(self,nums: List[int]) ->int: l, r =0,len(nums)-1while l < r: mid = (l+r)//2if nums[mid]> nums[mid+1]: r = midelse: l = mid+1return l
222. Count Complete Tree Nodes
Given a complete binary tree, count the number of nodes.
Can you solve in O(d) time
"""The node number in last level: 1 ~ 2^d-1Use 2 binary search, first search if idx exist in last level, second in 'exist function'Time O(2d)1. find depth of root2. return 1 if d == 03. Use 'exist' function to find if left idx exist4. Implement exist(idx, root, d)"""classSolution:defcountNodes(self,root: TreeNode) ->int:ifnot root:return0 d = self.getDepth(root)if d ==0:return1 left, right =1,2**dwhile left < right: mid = (left + right)//2if self.exist(mid, root, d): left = mid +1else: right = midreturn (2**d-1) + leftdefgetDepth(self,node): d =0while node.left: node = node.left d +=1return ddefexist(self,idx,node,d): left, right =0,2**d-1for _ inrange(d): mid = (left + right)//2if idx <= mid: node = node.left right = midelse: node = node.right left = midreturn node isnotNone