0311._Sparse_Matrix_Multiplication.md

311. Sparse Matrix Multiplication

难度: Medium

刷题内容

原题连接

• https://leetcode.com/problems/sparse-matrix-multiplication/description/

内容描述

Given two sparse matrices A and B, return the result of AB.

You may assume that A's column number is equal to B's row number.

Example:

Input:

A = [
  [ 1, 0, 0],
  [-1, 0, 3]
]

B = [
  [ 7, 0, 0 ],
  [ 0, 0, 0 ],
  [ 0, 0, 1 ]
]

Output:

     |  1 0 0 |   | 7 0 0 |   |  7 0 0 |
AB = | -1 0 3 | x | 0 0 0 | = | -7 0 3 |
                  | 0 0 1 |

解题方案

思路 1

直接搞一个完全的稀疏矩阵然后一个一个地去看，这样万一内存不够怎么办，所以我们基于A建一个非0的新list,然后对应的求出结果，这样操作数会少一些， 并且对内存的要求也没有那么高,借用一句话I think, in perspective of Big4, they would have a HUGE sparse dataset. And would like to process them in a machine. So memory does not fit without the Table representation of sparse matrix. And this is efficient since can be loaded into a one machine

class Solution(object):
    def multiply(self, A, B):
        """
        :type A: List[List[int]]
        :type B: List[List[int]]
        :rtype: List[List[int]]
        """
        m, n, nB = len(A), len(A[0]), len(B[0])
        res = [[0] * nB for i in range(m)]
        
        idxA = []
        for i in range(len(A)):
            tmp = []
            for j in range(len(A[0])):
                if A[i][j] != 0:
                    tmp.append(j)
            idxA.append(tmp)
        print(idxA)
        
        for i in range(len(idxA)):
            for j in idxA[i]:
                for k in range(nB):
                    res[i][k] += A[i][j] * B[j][k]
                
        return res