1492. The kth Factor of n
這一個題目可以直接用暴力法來解，搜尋空間為 1 到 n/2 + 1，時間複雜度為 $$O(n/2) \approx O(n)$$ 。
class Solution:
def kthFactor(self, n: int, k: int) -> int:
for i in range(1, n//2+1):
if n % i == 0:
k -= 1
if k == 0:
return i
return n if k == 1 else -1
不過既然是第 k 個項目，多數只要是找第 k 個項目的題目，都可以試著用 heap 來想。
class Solution:
def kthFactor(self, n: int, k: int) -> int:
heap = []
for x in range(1, int(n**0.5) + 1):
if n % x == 0:
heappush(heap, - x)
if len(heap) > k:
heappop(heap)
if x != n // x:
heappush(heap, - n // x)
if len(heap) > k:
heappop(heap)
return -heappop(heap) if k == len(heap) else -1