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

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

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
    }
}