zoukankan      html  css  js  c++  java
  • 【leetcode】1464. Maximum Product of Two Elements in an Array

    题目如下:

    Given the array of integers nums, you will choose two different indices i and j of that array. Return the maximum value of (nums[i]-1)*(nums[j]-1).

    Example 1:

    Input: nums = [3,4,5,2]
    Output: 12 
    Explanation: If you choose the indices i=1 and j=2 (indexed from 0), you will get the maximum value, that is, 
    (nums[1]-1)*(nums[2]-1) = (4-1)*(5-1) = 3*4 = 12.

    Example 2:

    Input: nums = [1,5,4,5]
    Output: 16
    Explanation: Choosing the indices i=1 and j=3 (indexed from 0), you will get the maximum value of (5-1)*(5-1) = 16.
    

    Example 3:

    Input: nums = [3,7]
    Output: 12

    Constraints:

    • 2 <= nums.length <= 500
    • 1 <= nums[i] <= 10^3

    解题思路:最直接的方法计算两个for循环。

    代码如下:

    class Solution(object):
        def maxProduct(self, nums):
            """
            :type nums: List[int]
            :rtype: int
            """
            res = 0
            for i in range(len(nums)):
                for j in range(len(nums)):
                    if i == j:continue
                    res = max(res,(nums[i]-1)*(nums[j]-1))
            return res
  • 相关阅读:
    完整的开源和商业软件平台
    免费开源的文件比较/合并工具
    Javascript面向对象基础
    Javascript面向对象基础
    引入外部js获取dom为null的问题
    闭包函数
    初识对象
    构造函数
    内置对象
    Math对象
  • 原文地址:https://www.cnblogs.com/seyjs/p/13176484.html
Copyright © 2011-2022 走看看