LeetCode（Sorting)1464. Maximum Product of Two Elements in an Array

╰半橙微兮° 2023-09-23 23:05 42阅读 0赞

# 1.问题 #

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

# 2. 解题思路 #

## 方法1： ##

> 1.定义变量max和secondMax的初始值为0  
> 2.for循环遍历，如果数组元素大于max，secondMax，max等于数组元素；如果secondMax小于数组元素，max大于secondMax元素，secondMax等于value  
> 3.计算result=(max-1)\*(secondMax-1);并返回result

## 方法2： ##

> 1.对nums的大小排序，nums.length-1\]-1是最大值索引，nums.length-2是最小值索引  
> 2. 计算(nums\[i\]-1)\*(nums\[j\]-1)的结果并返回

# 3. 代码 #

## 代码1： ##

class Solution {
        
        public int maxProduct(int[] nums) {
        
            int max=0;//1.定义变量max和secondMax的初始值为0
            int secondMax=0;
    
            for(int value:nums){
        //2.for循环遍历，如果数组元素大于max，secondMax，max等于数组元素；如果secondMax小于数组元素，max大于secondMax元素，secondMax等于value
                if(value>max){
        
                    secondMax= max;
                    max=value;
                }else if (secondMax<value&& secondMax<max){
        
                    secondMax=value;
                }
            }
            int result = (max-1)*(secondMax-1);//3.计算result=(max-1)*(secondMax-1);并返回result
            return result;
        }
    }

## 代码2： ##

class Solution {
        
        public int maxProduct(int[] nums) {
        
            Arrays.sort(nums);//1.对nums的大小排序，nums.length-1]-1是最大值索引，nums.length-2是最小值索引
            return (nums[nums.length-1]-1)*(nums[nums.length-2]-1);//2. 计算(nums[i]-1)*(nums[j]-1)的结果并返回
        }
    }