정답: (O)/(O)/(X)/(O)/(O)/(X)
def heapify(x):
"""Transform list into a heap, in-place, in O(len(x)) time."""
n = len(x)
# Transform bottom-up. The largest index there's any point to looking at
# is the largest with a child index in-range, so must have 2*i + 1 < n,
# or i < (n-1)/2. If n is even = 2*j, this is (2*j-1)/2 = j-1/2 so
# j-1 is the largest, which is n//2 - 1. If n is odd = 2*j+1, this is
# (2*j+1-1)/2 = j so j-1 is the largest, and that's again n//2-1.
for i in reversed(range(n//2)):
_siftup(x, i)