DeepSeek LeetCode 2040.两个有序数组的第 K 小乘积 Python3实现

```python
from typing import List
import bisect

class Solution:
def kthSmallestProduct(self, nums1: List[int], nums2: List[int], k: int) -> int:
# 统计乘积 <= mid 的个数
def count(mid: int) -> int:
cnt = 0
for a in nums1:
if a > 0:
# a * b <= mid => b <= floor(mid / a)
target = mid // a
cnt += bisect.bisect_right(nums2, target)
elif a < 0:
# a * b <= mid => b >= ceil(mid / a) = ceil(-mid / -a)
a_pos = -a
target = (-mid + a_pos - 1) // a_pos # 向上取整公式
cnt += len(nums2) - bisect.bisect_left(nums2, target)
else: # a == 0
if mid >= 0:
cnt += len(nums2)
return cnt

# 答案可能的最小值与最大值（四个角乘积的最值）
cand = [
nums1[0] * nums2[0],
nums1[0] * nums2[-1],
nums1[-1] * nums2[0],
nums1[-1] * nums2[-1],
]
left, right = min(cand), max(cand)

# 二分搜索答案
while left < right:
mid = (left + right) // 2
if count(mid) < k:
left = mid + 1
else:
right = mid
return left
```