350. Intersection of Two Arrays II

最新推荐文章于 2023-12-07 17:04:38 发布

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最新推荐文章于 2023-12-07 17:04:38 发布

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本文介绍了一种计算两个数组交集的算法实现，并讨论了不同情况下算法的优化方案。通过示例展示如何使用排序和双指针技巧高效地找出两个数组中的公共元素。

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

Example:
Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2, 2].

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?

翻译：给定两个数组，返回数组中数的交集。

class Solution{
public:
	vector<int> intersect(vector<int>& A,vector<int>& B)
	{
		int len1,len2;
		vector<int> result;
		sort(A.beign(),A.end());
		sort(B.beign(),B.end());
		len1 = A.size();
		len2 = B.size();
		while(i<len1 && j<len2)
		{
			if(A[i]==B[j])
			{
				res.push_back(A[i]);
				i++;
				j++;
			}
			else if(A[i]>B[j])
				j++;
			else
				i++;
		}
		return result;
	}
};