Python :compute two arrays‘ intersection

最新推荐文章于 2024-12-26 16:36:11 发布

原创最新推荐文章于 2024-12-26 16:36:11 发布 · 398 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#Python #Algorithms

Python 同时被 2 个专栏收录

2 篇文章

订阅专栏

Algorithms

2 篇文章

订阅专栏

本文介绍了一种计算两个数组交集的算法，通过实例展示了输入输出情况，并探讨了不同情况下算法的优化策略，包括已排序数组、大小差异显著的数组以及大规模数据处理。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Given two arrays, write a function to compute their intersection.

给定两个数组，编写一个函数来计算它们的交集。

EX1:

Input: nums1 = [1,2,2,1], nums2 = [2,2]
Output: [2,2]

EX2:

Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
Output: [4,9]

Note:

Each element in the result should appear as many times as it shows in both arrays.
The result can be in any order.
输出结果中每个元素出现的次数，应与元素在两个数组中出现的次数一致。
我们可以不考虑输出结果的顺序。

Follow up:

What if the given array is already sorted? How would you optimize your algorithm?
What if nums1's size is small compared to nums2's size? Which algorithm is better?
What if elements of nums2 are stored on disk, and the memory is limited such that you cannot load all elements into the memory at once?
如果给定的数组已经排好序呢？你将如何优化你的算法？
如果 nums1 的大小比 nums2 小很多，哪种方法更优？
如果 nums2 的元素存储在磁盘上，磁盘内存是有限的，并且你不能一次加载所有的元素到内存中，你该怎么办？

Sol One

If two arrays 's numbers arent too big, and for the big size of arrays, there is no doubt that the idea of #COUNTING-SORT will preform really well.

毫无疑问，如果对于数字较小，而规格大的数组，用#COUNTING-SORT类似的方法是可以表现得非常棒！

Here is the code:

class Solution(object):
    def intersect(self, nums1, nums2):
        k=10
        C=[]
        B=[]
        for i in range(k):
            C.append(0)
        for j in range(len(nums1)):
            C[nums1[j]-1]=C[nums1[j]-1]+1
        for j in range(len(nums2)):
            if C[nums2[j]-1]>0:
                B.append(nums2[j])
                C[nums2[j]-1]=C[nums2[j]-1]-1
        return B

Sol Two:

But when the number is bigger than 10**3, it becomes really too big to run.

Here s the general solution.

class Solution:
    def intersect(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: List[int]
        """
        record, result = {}, []
        for num in nums1:
            record[num] = record.get(num, 0) + 1
                
        for num in nums2:
            if num in record and record[num]:
                result.append(num)
                record[num] -= 1
        return result