题目描述

Given two sparse vectors, compute their dot product.

Implement class SparseVector:

• SparseVector(nums) Initializes the object with the vector nums
• dotProduct(vec) Compute the dot product between the instance of SparseVector and vec

A sparse vector is a vector that has mostly zero values, you should store the sparse vector efficiently and compute the dot product between two SparseVector.

Follow up: What if only one of the vectors is sparse?

样例

Input: nums1 = [1,0,0,2,3], nums2 = [0,3,0,4,0]
Output: 8
Explanation: v1 = SparseVector(nums1) , v2 = SparseVector(nums2)
v1.dotProduct(v2) = 1*0 + 0*3 + 0*0 + 2*4 + 3*0 = 8

算法1 (Index-Value Pairs模拟)

时间复杂度: $O(n)$ for creating non-zero index-value pairs, $O(l_1 + l_2)$ for dot product

空间复杂度: $O(l_1)$ and $O(1)$

C++ 代码

typedef pair<int, int> PII;
class SparseVector {
public:
    vector<PII> v;
    SparseVector(vector<int> &nums) {
        for (int i = 0; i < nums.size(); i++)
            if (nums[i])
                v.push_back({i, nums[i]});
    }

    // Return the dotProduct of two sparse vectors
    int dotProduct(SparseVector& vec) {
        int res = 0;
        for (int i = 0, j = 0; i < v.size() && j < vec.v.size(); i++, j++) {
            if (v[i].first < vec.v[j].first) j--;
            else if (v[i].first > vec.v[j].first) i--;
            else res += v[i].second * vec.v[j].second;
        }
        return res;
    }
};

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

算法2 (Index-Value Pairs, 二分)

For the follow-up question, if the length of one sparse vector’s non-zero element is much greater than the other one’s, we could use binary search on the long sparse vector.

时间复杂度: $O(min(l_1, l_2) * \log (max(l_1, l_2)))$

空间复杂度: 同上

C++ 代码

typedef pair<int, int> PII;
class SparseVector {
public:
    vector<PII> v;
    SparseVector(vector<int> &nums) {
        for (int i = 0; i < nums.size(); i++)
            if (nums[i])
                v.push_back({i, nums[i]});
    }

    // Return the dotProduct of two sparse vectors
    int dotProduct(SparseVector& vec) {
        int res = 0;
        vector<PII>& v_short = vec.v;
        vector<PII>& v_long = v;
        if (v_short.size() > v_long.size())
            swap(v_short, v_long);
        for (auto &p: v_short) {
            int idx = lower_bound(v_long.begin(), v_long.end(), make_pair(p.first, - 1)) - v_long.begin();
            if (idx != v_long.size() && v_long[idx].first == p.first)
                res += v_long[idx].second * p.second;
        }
        return res;
    }
};

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