唉 有想到binary search 但我還是偷看了一下答案 現在連寫binary search也要想半天 我又鬱鬱了 def kthSmallestProduct(self, nums1: List[int], nums2: List[int], k: int) -> int: def count(x): # number of products which are less than or equal to x total_cnt = 0 for num1 in nums1: if num1>0: total_cnt += bisect_right(nums2, x/num1) elif num1==0: total_cnt += len(nums2) if x>=0 else 0 else: total_cnt += (len(nums2)-bisect_left(nums2, x/num1)) return total_cnt l, r = -10**10, 10**10 while l<r: mid = (l+r)//2 if count(mid)<k: l = mid+1 else: r = mid return l -- ※ 發信站: 批踢踢實業坊(pttsite.org.tw), 來自: 220.132.58.28 (臺灣) ※ 文章網址: https://pttsite.org.tw/Marginalman/M.1750862662.A.E0A

推 sustainer123: 大師 06/25 22:51