📅  最后修改于: 2023-12-03 15:40:14.450000             🧑  作者: Mango
本题的目标是在两个整数数组中寻找唯一对,使得这对元素之间的差值的绝对值最大化。这个问题可以被解决为一个经典问题,即寻找两个有序集合中的最大的差值,而且有多种解决方法。
暴力搜索是一种最直接的解决方式。它的复杂度为 $O(n^2)$,意味着在较大的数据集上运行速度很慢,但在小数据集上可以提供可接受的性能表现。
def find_max_diff(arr1, arr2):
max_diff = float('-inf')
max_pair = None
for x in arr1:
for y in arr2:
diff = abs(x - y)
if diff > max_diff:
max_diff = diff
max_pair = (x, y)
return max_pair
排序后搜索是一种比较高效的解决方法,虽然需要较大的内存空间,但时间复杂度为 $O(n \log n)$,在较大的数据集上表现更加优异。
def find_max_diff(arr1, arr2):
arr1.sort()
arr2.sort()
i, j = 0, 0
max_diff = float('-inf')
max_pair = None
while i < len(arr1) and j < len(arr2):
diff = abs(arr1[i] - arr2[j])
if diff > max_diff:
max_diff = diff
max_pair = (arr1[i], arr2[j])
if arr1[i] < arr2[j]:
i += 1
else:
j += 1
return max_pair
二分查找是一种时间复杂度为 $O(n \log n)$ 的解决方法,它利用两个数组均为有序的事实,基于二分查找的思想,在一个列表中查找另一个列表的元素,以此得到两个数组中的唯一对。
from bisect import bisect_left
def binary_search(arr, val):
i = bisect_left(arr, val)
if i != len(arr) and arr[i] == val:
return val
elif i != 0 and abs(arr[i-1]-val) < abs(arr[i]-val):
return arr[i-1]
else:
return arr[i]
def find_max_diff(arr1, arr2):
arr2_set = set(arr2)
max_diff = float('-inf')
max_pair = None
for val in arr1:
floor_val = binary_search(arr2, val)
if floor_val is not None:
diff = abs(val - floor_val)
if diff > max_diff:
max_diff = diff
max_pair = (val, floor_val)
return max_pair
哈希表是以空间换时间的解决方案,解决了内存空间的需求,成为最理想的方法。时间复杂度为 $O(n)$。
def find_max_diff(arr1, arr2):
max_diff = float('-inf')
max_pair = None
diff_map = {}
for i in range(len(arr2)):
diff_map[arr2[i]] = i
for val in arr1:
if (val + max_diff) in diff_map:
max_pair = (val, (val + max_diff))
if (val - max_diff) in diff_map:
max_pair = (val, (val - max_diff))
if val in diff_map:
max_diff = max(max_diff, diff_map[val] - val)
return max_pair
以上是求解本题的常用方法。根据实际需求,选择适合的方法进行实现。