【leetcode】1464. Maximum Product of Two Elements in an Array

标签：pre   diff   you   nts   res   maximum   list   osi   length   

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

【leetcode】1464. Maximum Product of Two Elements in an Array

标签：pre   diff   you   nts   res   maximum   list   osi   length   

原文地址：https://www.cnblogs.com/seyjs/p/13176484.html