8/7 每日一題(找出最大成積)

 Given an integer array nums, find three numbers whose product is maximum and return the maximum product.

 

Example 1:

Input: nums = [1,2,3]
Output: 6

Example 2:

Input: nums = [1,2,3,4]
Output: 24

Example 3:

Input: nums = [-1,-2,-3]
Output: -6

 

Constraints:

• 3 <= nums.length <= 104
• -1000 <= nums[i] <= 1000

class Solution:

    def maximumProduct(self, nums: List[int]) -> int:        

        nums.sort()

       

        b=nums[0]*nums[1]*nums[-1]

        c=nums[-1]*nums[-2]*nums[-3]

       

       

        return max(b,c)

       

       

     

---------------參考解答

class Solution:
    def maximumProduct(self, nums: List[int]) -> int:
        pos=heapq.nlargest(3,nums)     #找出nums中3個最大值加入pos
        neg=heapq.nsmallest(2,nums)  #找出nums中2個最小值加入neg

        return max(neg[0]*neg[1]*pos[0],pos[0]*pos[1]*pos[2])