[LeetCode] 628. Maximum Product of Three Numbers 三个数字的最大乘积

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.

给一个整数数组，找出乘积最大的3个数，返回这3个数组。

解法：这题的难点主要是要考虑到负数和0的情况。最大乘积可能是最大的3个正数相乘或者是最小的2个负数和最大的1个正数相乘。

Java:

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public int maximumProduct(int[] nums) {         int max1 = Integer.MIN_VALUE, max2 = Integer.MIN_VALUE, max3 = Integer.MIN_VALUE, min1 = Integer.MAX_VALUE, min2 = Integer.MAX_VALUE;         for (int n : nums) {             if (n > max1) {                 max3 = max2;                 max2 = max1;                 max1 = n;             } else if (n > max2) {                 max3 = max2;                 max2 = n;             } else if (n > max3) {                 max3 = n;             }               if (n < min1) {                 min2 = min1;                 min1 = n;             } else if (n < min2) {                 min2 = n;             }         }         return Math.max(max1*max2*max3, max1*min1*min2);     }

Python:

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# Time:  O(n) # Space: O(1) class Solution(object):     def maximumProduct(self, nums):         """         :type nums: List[int]         :rtype: int         """         min1, min2 = float("inf"), float("inf")         max1, max2, max3 = float("-inf"), float("-inf"), float("-inf")           for n in nums:             if n <= min1:                 min2 = min1                 min1 = n             elif n <= min2:                 min2 = n               if n >= max1:                 max3 = max2                 max2 = max1                 max1 = n             elif n >= max2:                 max3 = max2                 max2 = n             elif n >= max3:                 max3 = n           return max(min1 * min2 * max1, max1 * max2 * max3)

Python:

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

Python:

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def maximumProduct(self, nums):         a, b = heapq.nlargest(3, nums), heapq.nsmallest(2, nums)         return max(a[0] * a[1] * a[2], b[0] * b[1] * a[0])

C++:

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class Solution { public:     int maximumProduct(vector<int>& nums) {         int mx1 = INT_MIN, mx2 = INT_MIN, mx3 = INT_MIN;         int mn1 = INT_MAX, mn2 = INT_MAX;         for (int num : nums) {             if (num > mx1) {                 mx3 = mx2; mx2 = mx1; mx1 = num;             } else if (num > mx2) {                 mx3 = mx2; mx2 = num;             } else if (num > mx3) {                 mx3 = num;             }             if (num < mn1) {                 mn2 = mn1; mn1 = num;             } else if (num < mn2) {                 mn2 = num;             }         }         return max(mx1 * mx2 * mx3, mx1 * mn1 * mn2);     } };

　

类似题目：

[LeetCode] 152. Maximum Product Subarray 求最大子数组乘积

　

 

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