这是 O（log n）的算法存在。我们采用分而治之。

An O(log n) algorithm exists. We use divide-and-conquer.

find_peak(lo,hi):
  mid = (lo+hi)/2
  if A[mid] >= A[mid-1], A[mid+1] return mid
  if A[mid] < A[mid-1] 
    return find_peak(lo,mid-1) // a peak must exists in A[1..mid-1]
  if A[mid] < A[mid+1]
    return find_peak(mid+1,hi) // a peak must exists in A[mid+1..hi]