LeetCode 311. Sparse Matrix Multiplication

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本文介绍了一种针对稀疏矩阵乘法的高效算法实现。通过预先构建B矩阵的映射表来减少不必要的乘法运算，特别适用于大量元素为零的矩阵。文中提供了两种实现方式并比较了其效率。

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#include <vector>
#include <iostream>
#include <map>
using namespace std;
/*
  Given two sparse matrics A and B, return the result of AB.
  You may assume that A's column number is equal to B's row number.
  For example:
  A = [
  [ 1, 0, 0],
  [-1, 0, 3]
]

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


     |  1 0 0 |   | 7 0 0 |   |  7 0 0 |
AB = | -1 0 3 | x | 0 0 0 | = | -7 0 3 |
                  | 0 0 1 |
*/
vector< vector<int> > matrixMultiply(vector< vector<int> >& A, vector< vector<int> >& B) {
  int m = A.size(), n = A[0].size(), k = B[0].size();
  vector< vector<int> > res(m, vector<int>(k, 0));
  for(int i = 0; i < m; ++i) {
    for(int j = 0; j < n; ++j) {
      if(A[i][j] != 0) {
        for(int c = 0; c < k; ++c) {
          if(B[j][c] != 0) {
              res[i][c] = A[i][j] * B[j][c];
          }
        }
      }
    }
  }
  return res;
}
vector< vector<int> > matrixMultiplyII(vector< vector<int> >& A, vector< vector<int> >& B) {
  int m = A.size(), n = A[0].size(), l = B[0].size();
  vector< vector<int> > res(m, vector<int>(l, 0));
  map<int, map<int, int> > tableB;
  for(int i = 0; i < n; ++i) {
    for(int j = 0; j < l; j++) {
      map<int, int> tmp;
      tmp.insert({j, B[i][j]});
      if(B[i][j] != 0) {tableB[i] = tmp;}
    }
  }
  for(int i = 0; i < m; ++i) {
    for(int j = 0; j < n; ++j) {
      if(A[i][j] != 0) {
        map<int, int> tmp = tableB[j];
        auto iter = tmp.begin();
        while(iter != tmp.end()) {
          res[i][iter->first] += A[i][j] * iter->second;
          iter++;
        }
      }
    }
  }
  return res;
}

int main(void) {
  vector< vector<int> > A{
    {1, 0, 0},
    {-1, 0, 3}};
  vector< vector<int> > B{
    {7, 0, 0},
    {0, 0, 0},
    {0, 0, 1}};
  vector< vector<int> > res = matrixMultiplyII(A, B);
  for(int i = 0; i < res.size(); ++i) {
    for(int j = 0; j < res[i].size(); ++j) {
      cout << res[i][j] << " ";
    }
    cout << endl;
  }
}