[LeetCode] 628. Maximum Product of Three Numbers_Easy

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

Example 1:

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

Example 2:

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

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.

这个题目就是需要注意负数的情况即可, 我们如果用排序的话, 比较max(nums[0]*nums[1]*nums[-1], nums[-1]*nums[-2]*nums[-3])即可. T: O(nlgn)

要T: O(n) 的话就需要得到min1, min2, max1, max2, max3同理比较max(min1 * min2 * max1, max1 * max2 * max3)即可.

Code

1) T: O(nlgn)

class Solution:
    def maxProduct3nums(self, nums):
        nums.sort()
        return max(nums[0]*nums[1]*nums[-1], nums[-1]*nums[-2]*nums[-3])

2) T: O(n)

class Solution:
    def maxProduct3nums(self, nums):
        ans = [1001]*2 + [-1001]*3  # min1, min2, max1, max2, max3
        for num in nums:
            if num > ans[-1]:
                ans[-1], ans[-2], ans[-3] = num, ans[-1], ans[-2]
            elif num > ans[-2]:
                ans[-2], ans[-3] = num, ans[-2]
            elif num > ans[-3]:
                ans[-3] = num
            if num < ans[0]:
                ans[0], ans[1] = num, ans[0]
            elif num < ans[1]:
                ans[1] = num
        return max(ans[0]*ans[1]*ans[-1], ans[-1]*ans[-2]*ans[-3])

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