LeetCode 1570. 两个稀疏向量的点积（哈希）

示例 1：
输入：nums1 = [1,0,0,2,3], nums2 = [0,3,0,4,0]
输出：8
解释：v1 = SparseVector(nums1) , v2 = SparseVector(nums2)
v1.dotProduct(v2) = 1*0 + 0*3 + 0*0 + 2*4 + 3*0 = 8

示例 2：
输入：nums1 = [0,1,0,0,0], nums2 = [0,0,0,0,2]
输出：0
解释：v1 = SparseVector(nums1) , v2 = SparseVector(nums2)
v1.dotProduct(v2) = 0*0 + 1*0 + 0*0 + 0*0 + 0*2 = 0

示例 3：
输入：nums1 = [0,1,0,0,2,0,0], nums2 = [1,0,0,0,3,0,4]
输出：6
 
提示：
n == nums1.length == nums2.length
1 <= n <= 10^5
0 <= nums1[i], nums2[i] <= 100

class SparseVector {
public:
    unordered_map<int,int> m;
    int size = 0;
    SparseVector(vector<int> &nums) {
        size = nums.size();
        for(int i = 0; i < nums.size(); ++i)
        {
            if(nums[i])
                m[i] = nums[i];
        }
    }
    
    // Return the dotProduct of two sparse vectors
    int dotProduct(SparseVector& vec) {
        int ans = 0;
        for(auto& kv : vec.m)
        {
            if(m.find(kv.first) != m.end())
            {
                ans += m[kv.first]*kv.second;
            }
        }
        return ans;
    }
};

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