[Swift]LeetCode311. 稀疏矩阵相乘 $ Sparse Matrix Multiplication

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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:

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 |

---

给定两个稀疏矩阵A和B，返回AB的结果。 

您可以假定A的列号等于B的行号。

例子：

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 |

---

Solution:

class Solution {
    func multiply(_ A:inout [[Int]],_ B:inout [[Int]]) ->[[Int]] {
        var res:[[Int]] = [[Int]](repeating:[Int](repeating:0,count:B[0].count),count:A.count)
        for i in 0..<A.count
        {
            for k in 0..<A[0].count
            {
                if A[i][k] != 0
                {
                    for j in 0..<B[0].count
                    {
                        if B[k][j] != 0
                        {
                            res[i][j] += A[i][k] * B[k][j]
                        }
                    }
                }
            }
        }
        return res            
    }
}

---

Solution:

class Solution {
    func multiply(_ A:inout [[Int]],_ B:inout [[Int]]) ->[[Int]] {
        var res:[[Int]] = [[Int]](repeating:[Int](repeating:0,count:B[0].count),count:A.count)
        var v:[[(Int,Int)]] = [[(Int,Int)]](repeating:[(Int,Int)](),count:A.count)
        for i in 0..<A.count
        {
            for k in 0..<A[i].count
            {
                if A[i][k] != 0
                {
                    v[i].append((k, A[i][k]))
                }
            }
        }
        for i in 0..<A.count
        {
            for k in 0..<v[i].count
            {
                var col:Int = v[i][k].0
                var val:Int = v[i][k].1
                for j in 0..<B[0].count
                {
                    res[i][j] += val * B[col][j]
                }
            }
        }
        return res            
    }
}

点击：Playground测试

var A:[[Int]] = [[ 1, 0, 0],[-1, 0, 3]]
var B:[[Int]] = [[ 7, 0, 0 ],[ 0, 0, 0 ],[ 0, 0, 1 ]]
let sol = Solution()
print(sol.multiply(&A,&B))
//Print [[7, 0, 0], [-7, 0, 3]]