Table of Contents
一、中文版
给你一个整数数组 nums
,请你选择数组的两个不同下标 i
和 j
,使 (nums[i]-1)*(nums[j]-1)
取得最大值。
请你计算并返回该式的最大值。
示例 1:
输入:nums = [3,4,5,2] 输出:12 解释:如果选择下标 i=1 和 j=2(下标从 0 开始),则可以获得最大值,(nums[1]-1)*(nums[2]-1) = (4-1)*(5-1) = 3*4 = 12 。
示例 2:
输入:nums = [1,5,4,5] 输出:16 解释:选择下标 i=1 和 j=3(下标从 0 开始),则可以获得最大值 (5-1)*(5-1) = 16 。
示例 3:
输入:nums = [3,7] 输出:12
提示:
2 <= nums.length <= 500
1 <= nums[i] <= 10^3
二、英文版
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
三、My answer
class Solution:
def maxProduct(self, nums: List[int]) -> int:
res = 0
for i in range(len(nums)):
for j in range(1,len(nums)):
if i != j and (nums[i]-1) * (nums[j]-1) > res:
res = (nums[i]-1) * (nums[j]-1)
return res
四、解题报告
直接遍历不会超时,所以可以双重 for 循环。
class Solution:
def maxProduct(self, nums: List[int]) -> int:
nums.sort()
return (nums[-1]-1)*(nums[-2]-1)