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

Note:

1. The length of the given array will be in range [3,104] and all elements are in the range [-1000, 1000].
2. Multiplication of any three numbers in the input won‘t exceed the range of 32-bit signed integer.

原題地址: Maximum Product of Three Numbers

難度: Easy

題意: 找出相乘最大的三個數

思路:

因為數字有正有負,因此相乘最大的三個數分為兩種情況:

(1)最大的三個正數

(2)最小的兩個負數以及一個最大的正數

代碼:

class Solution(object):
    def maximumProduct(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        a = b = c = None
        d 
 
= e = 0x7FFFFFFF
        for i in range(len(nums)):     
            if nums[i] >= a:           # 找出最大的三個數
                a, b, c = nums[i], a, b
            elif nums[i] >= b:
                b, c = nums[i], b
            elif nums[i] >= c:
                c = nums[i]

            if nums[i] <= e:　　　　　　# 找出最小的兩個數　
 
　　　
                d, e = e, nums[i]
            elif nums[i] <= d:
                d = nums[i]

        max_val = 0

#         if a > 0 and b > 0 and c > 0:
#             max_val = max(max_val, a * b * c)

#         if a > 0 and d < 0 and e < 0:
#             max_val = max(max_val, a * d * e)

#         if a < 0 and b < 0 and c < 0:
#             max_val = a * b * c
        max_val = max(a*b*c, a*d*e)
        return max_val

時間復雜度: O(n)

空間復雜度: O(1)

628. Maximum Product of Three Numbers@python