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],[-1,3] ] B = [ [ 7,0 ],[ 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 = [ [ 1,1 ] ] | 1 0 0 | | 7 0 0 | | 7 0 0 | AB = | -1 0 3 | x | 0 0 0 | = | -7 0 3 | | 0 0 1 |
Solution:
1 class Solution { 2 func multiply(_ A:inout [[Int]],_ B:inout [[Int]]) ->[[Int]] { 3 var res:[[Int]] = [[Int]](repeating:[Int](repeating:0,count:B[0].count),count:A.count) 4 for i in 0..<A.count 5 { 6 for k in 0..<A[0].count 7 { 8 if A[i][k] != 0 9 { 10 for j in 0..<B[0].count 11 { 12 if B[k][j] != 0 13 { 14 res[i][j] += A[i][k] * B[k][j] 15 } 16 } 17 } 18 } 19 } 20 return res 21 } 22 }
Solution:
1 class Solution { 2 func multiply(_ A:inout [[Int]],count:A.count) 4 var v:[[(Int,Int)]] = [[(Int,Int)]](repeating:[(Int,Int)](),count:A.count) 5 for i in 0..<A.count 6 { 7 for k in 0..<A[i].count 8 { 9 if A[i][k] != 0 10 { 11 v[i].append((k,A[i][k])) 12 } 13 } 14 } 15 for i in 0..<A.count 16 { 17 for k in 0..<v[i].count 18 { 19 var col:Int = v[i][k].0 20 var val:Int = v[i][k].1 21 for j in 0..<B[0].count 22 { 23 res[i][j] += val * B[col][j] 24 } 25 } 26 } 27 return res 28 } 29 }
点击:Playground测试
1 var A:[[Int]] = [[ 1,0,0],[-1,3]] 2 var B:[[Int]] = [[ 7,0 ],[ 0,1 ]] 3 let sol = Solution() 4 print(sol.multiply(&A,&B)) 5 //Print [[7,[-7,3]]
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