本文正在参加「Python主题月」，详情查看 活动链接
描述
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.
Note:
4 <= nums.length <= 10^4
1 <= nums[i] <= 10^4
解析
根据题意，就是给出一个整数数组 nums ，然后从 nums 中找出四个数字 a 、b 、c 、d ，使得 (a * b) - (c * d) 的结果是最大的，然后返回这个最大的值。这里用到的是内置的函数，因为这种两两相乘然后做差值的运算，肯定是要 a 和 b 用最大的数， c 和 d 用最小的数，这样结果才能最大，直接对 nums 进行排序，然后取最后的两个作为 a 和 b ，取最开始的两个数作为 c 和 d ，然后计算 (a * b) - (c * d) 得到的值即为结果。
解答
class Solution(object):
def maxProductDifference(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
nums.sort()
return nums[-1]*nums[-2]-nums[0]*nums[1]
运行结果
Runtime: 140 ms, faster than 77.76% of Python online submissions for Maximum Product Difference Between Two Pairs.
Memory Usage: 14.5 MB, less than 29.60% of Python online submissions for Maximum Product Difference Between Two Pairs.
解析
上面使用内置函数的解法虽然思路简单，操作也方便，但是不推荐。我们可以直接使用遍历列表的方法，一次梭哈，直接用 a 和 b 表示两个较大的数，c 和 d 表示两个较小的数，然后通过和 nums 中的每个元素进行比较，更新他们的值，遍历结束计算 a * b - c * d 即为答案。
解答
class Solution(object):
def maxProductDifference(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
a = b = float('-inf')
c = d = float('inf')
for n in nums:
if n <= c:
c, d, = n, c
elif n < d:
d = n
if n >= a:
a, b = n, a
elif n > b:
b = n
return a * b - c * d
运行结果
Runtime: 140 ms, faster than 77.76% of Python online submissions for Maximum Product Difference Between Two Pairs.
Memory Usage: 14.4 MB, less than 76.29% of Python online submissions for Maximum Product Difference Between Two Pairs.
解析
当然了我们仍然可以使用最无脑的暴力解决办法，但是性能太差了，四重循环求最大值，不推荐。
解析
class Solution(object):
def maxProductDifference(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
result = 0
for i,a in enumerate(nums):
for j,b in enumerate(nums):
for k,c in enumerate(nums):
for l,d in enumerate(nums):
if len(set([i,j,k,l])) == len([i,j,k,l]):
result = max(result, a*b-c*d)
return result
运行结果
Time Limit Exceeded
原题链接：leetcode.com/problems/ma…
您的支持是我最大的动力
