1570. Dot Product of Two Sparse Vectors

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?

Example 1:

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

Example 2:

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

Example 3:

Input: nums1 = [0,1,0,0,2,0,0], nums2 = [1,0,0,0,3,0,4]
Output: 6

Constraints:

• n == nums1.length == nums2.length

• 1 <= n <= 10^5

分析

可能是稀疏矩阵，所以只存非空数和索引

class SparseVector:
    def __init__(self, nums: List[int]):
        self.nums = [(i,v) for i,v in enumerate(nums) if v]

    # Return the dotProduct of two sparse vectors
    def dotProduct(self, vec: 'SparseVector') -> int:
        res = 0
        i = j = 0
        while i < len(self.nums) and j < len(vec.nums):
            if self.nums[i][0] == vec.nums[j][0]:
                res += self.nums[i][1] * vec.nums[j][1]
                j += 1
                i += 1
            elif self.nums[i][0] < vec.nums[j][0]:
                i += 1
            else:
                j += 1
        return res
            
# Your SparseVector object will be instantiated and called as such:
# v1 = SparseVector(nums1)
# v2 = SparseVector(nums2)
# ans = v1.dotProduct(v2)