LeetCode（Sorting)1913. Maximum Product Difference Between Two Pairs

ゝ一纸荒年。 2023-09-23 23:11 93阅读 0赞

# 1.问题 #

The product difference between two pairs (a, b) and (c, d) is defined as (a \* b) - (c \* d).

For example, the product difference between (5, 6) and (2, 7) is (5 \* 6) - (2 \* 7) = 16.  
Given an integer array nums, choose four distinct indices w, x, y, and z such that the product difference between pairs (nums\[w\], nums\[x\]) and (nums\[y\], nums\[z\]) is maximized.

Return the maximum such product difference.

Example 1:

> Input: nums = \[5,6,2,7,4\]  
> Output: 34  
> Explanation: We can choose indices 1 and 3 for the first pair (6, 7) and indices 2 and 4 for the second pair (2, 4).  
> The product difference is (6 \* 7) - (2 \* 4) = 34.

Example 2:

> Input: nums = \[4,2,5,9,7,4,8\]  
> Output: 64  
> Explanation: We can choose indices 3 and 6 for the first pair (9, 8) and indices 1 and 5 for the second pair (2, 4).  
> The product difference is (9 \* 8) - (2 \* 4) = 64.

Constraints:

*  4 <= nums.length <= 104
 *  1 <= nums\[i\] <= 104

# 2. 解题思路 #

## 方法1： ##

> 1.nums升序排序  
> 2.nums\[w\]\*nums\[x\]) -nums\[y\]\*nums\[z\] 的最大值，排序后数组的倒数前两个数值乘积减去排序后数组的正数前两个数值乘积  
> 3.返回result结果

## 方法2： ##

> 设定变量，for循环或for each遍历， 比较最大，次最大， 最小，次最小，之后通过运算求值

# 3. 代码 #

## 代码1： ##

class Solution {
        
        public int maxProductDifference(int[] nums) {
        
             Arrays.sort(nums);//1.nums升序排序
             int result=(nums[nums.length-1]*nums[nums.length-2])-(nums[0]*nums[1]);//2.nums[w]*nums[x]) -nums[y]*nums[z] 的最大值，排序后数组的倒数前两个数值乘积减去排序后数组的正数前两个数值乘积
            }
            return result;//3.返回result结果
            
        }
    }

## 代码2： ##

class Solution {
        
        public int maxProductDifference(int[] nums) {
        
            int max1 = Integer.MIN_VALUE;
            int max2 = Integer.MIN_VALUE;
            
            int min1 = Integer.MAX_VALUE;
            int min2 = Integer.MAX_VALUE;
            
            for(int val : nums){
        
                if(val > max1){
        
                    max2 = max1;
                    max1 = val;
                } else if(val > max2){
        
                    max2 = val;
                }
                
                if(val < min1){
        
                    min2 = min1;
                    min1 = val;
                } else if(val < min2){
        
                    min2 = val;
                }
            }
            
            return (max1 * max2) - (min1 * min2);
        }
    }