350. Intersection of Two Arrays II

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最新推荐文章于 2023-12-07 17:04:38 发布

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本文介绍两种计算两个数组交集的方法：暴力破解法和哈希表法。暴力破解法通过双重循环匹配元素并移除已匹配项；哈希表法则利用哈希表统计元素频率，再遍历第二个数组进行匹配。

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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?

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【思路一】暴力破解

class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        vector <int>  v;
        int n = nums2.size();
        for(int i =0;i<n;i++){
            for(int j=0;j<nums1.size();j++){
                if(nums2[i]==nums1[j]){
                    v.push_back(nums2[i]);
                    nums1.erase(nums1.begin()+j);
                    break;
                }
                
            }
        }
        return v;
    }
};

【思路二】hash table

class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        if(nums1.size()==0 || nums2.size()==0) return vector<int>();
        vector<int> result;
        unordered_map<int, int> hash;
        for(auto val: nums1) hash[val]++;
        for(auto val: nums2)
        {
            if(hash.count(val) && hash[val]>0) result.push_back(val);
            hash[val]--;
        }
        return result;
    }
};