使用哈希表记录每个稀疏向量的下标和值

然后找出哈希表中的公共项相乘即可

时间复杂度$O(n)$

class SparseVector {
public:
    unordered_map<int,int>hash;
    SparseVector(vector<int> &nums) {
        for(int i=0;i<nums.size();i++)
            hash[i]=nums[i];
    }

    // Return the dotProduct of two sparse vectors
    int dotProduct(SparseVector& vec) {
        int res=0;
        unordered_map<int,int>hash2=vec.hash;
        for(auto [c,v]:hash)
        {
            if(hash2.count(c))
            {
                res+=v*hash2[c];
            }
        }
        return res;
    }
};

// Your SparseVector object will be instantiated and called as such:
// SparseVector v1(nums1);
// SparseVector v2(nums2);
// int ans = v1.dotProduct(v2);