The length of the given array will be in range [3,104] and all elements are in the range [-1000, 1000]. Multiplication of any three numbers in the input won't exceed the range of 32-bit signed integer. 分析:题目要求给一个数组,求出这个数组三个数字乘积的最大值。乍一看比较简单,就是...
The length of the given array will be in range [3,104] and all elements are in the range [-1000, 1000]. Multiplication of any three numbers in the input won't exceed the range of 32-bit signed integer. 这道题博主刚开始看的时候,心想直接排序,然后最后三个数字相乘不就完了,心想不会这么...
The length of the given array will be in range [3,104] and all elements are in the range [-1000, 1000]. Multiplication of any three numbers in the input won't exceed the range of 32-bit signed integer. 这个题目本质上是一个数学题,需要细心观察才可以得到答案,考虑输入数据分布的几种情形:...
class Solution{public:intmaximumProduct(vector<int>&nums){intmax1=-1000,max2=-1000,max3=-1000,min1=1000,min2=1000;for(autonum:nums){if(num>max1){max3=max2;max2=max1;max1=num;}elseif(num>max2){max3=max2;max2=num;}elseif(num>max3)max3=num;if(num<min1){min2=min1;min1=...
Multiplication of any three numbers in the input won’t exceed the range of 32-bit signed integer. 分析 题目的意思是:找出一个数组中三个数的乘积最大。 考虑正负数的情况。 如果全都是正数相乘比较大,就取三个最大值相乘即可。 如果负数的绝对值比较大,我们可以取绝对值最大的两个负数参与相乘,最后比...
funcmaximumProduct(_nums:[Int])->Int{ifnums.count==3{returnnums[0]*nums[1]*nums[2]}vartmp=nums.sorted()varm=Int.miniftmp[tmp.count-3]>=0&&tmp[tmp.count-2]>=0&&tmp[tmp.count-1]>=0{m=max(m,tmp[tmp.count-3]*tmp[tmp.count-2]*tmp[tmp.count-1])}iftmp.first!<=0&&tmp[...
leetcode 628. Maximum Product of Three Numbers(三个数的最大乘积)--Java题解,程序员大本营,技术文章内容聚合第一站。
简介:Given an integer array, find three numbers whose product is maximum and output the maximum product. Given an integer array, find three numbers whose product is maximum and output the maximum product. Example 1: Input: [1,2,3]
Can you solve this real interview question? Maximum Product of Three Numbers - 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: n
# @lc app=leetcode.cn id=628 lang=python # # [628] 三个数的最大乘积 # # @lc code=start class Solution(object): def maximumProduct(self, nums): """ :type nums: List[int] :rtype: int """ negative_nums = [num for num in nums if num < 0] positive_nums = [num for num ...