LeetCode350. Intersection of Two Arrays II

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本文深入探讨了两个数组交集的算法实现，通过实例展示了如何计算两个数组的交集，并详细解释了当输入数组已排序、大小不一或部分元素存储在磁盘上的情况下，如何优化算法。文章提供了一种基于双指针的解决方案，能够有效处理重复元素，适用于多种场景。

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

Example 1:

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

Example 2:

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.

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?

思路:与上一题相比,稍微复杂一些, 需要注意在两个列表中分别存在的个数.

可以将两个列表进行排序,然后用两个指针,同时从头开始比较,这样遇到相同元素时,就可以统计出现次数.

class Solution:
    def intersect(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: List[int]
        """
        nums1.sort()
        nums2.sort()
        same=[]
        a,b=0,0
        while a<len(nums1) and b<len(nums2):
            if nums1[a] == nums2[b]:
                same.append(nums1[a])
                a += 1
                b += 1
            elif nums1[a]<nums2[b]:
                a += 1
            else:
                b += 1
            
        return same