[350]Intersection of Two Arrays II

最新推荐文章于 2020-10-16 02:55:14 发布

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本文探讨了两个数组交集的算法实现，通过示例展示如何返回两个数组中重复出现的所有元素，并讨论了不同场景下的优化策略，如已排序数组及大规模数据处理。

【题目描述】

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?

【题目思路】

感觉比349更简单，因为这次纯粹是求两个数集的交集，不用考虑是否元素重复，follow up里的第三个子问题还是比较有意义的，从discuss里看到的回答，感觉不错

Q：

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?

A：

If only nums2 cannot fit in memory, put all elements of nums1 into a HashMap, read chunks of array that fit into the memory, and record the intersections.
If both nums1 and nums2 are so huge that neither fit into the memory, sort them individually (external sort), then read 2 elements from each array at a time in memory, record intersections.

另外每次取两个数可能太慢了，毕竟数集可能很大，可以每次取2g的数据进行处理，可以相对快点。

【代码】

class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        int len1=nums1.size();  
        int len2=nums2.size();  
        vector<int> ans;  
        if(len1==0||len2==0) return ans;  
        int i=0,j=0;  
        sort(nums1.begin(),nums1.end());  
        sort(nums2.begin(),nums2.end());  
        while(i<len1&&j<len2){  
           // while(i<len1-1 && nums1[i]==nums1[i+1]) i++;  
            //while(j<len2-1 && nums2[j]==nums2[j+1]) j++;  
            if(nums1[i]==nums2[j]){  
                ans.push_back(nums1[i]);  
                i++;  
                j++;  
            }  
            else{  
                if(nums1[i]<nums2[j]) i++;  
                else j++;  
            }  
        }  
        return ans;  
    }
};