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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 |
1 public class Solution { 2 public int[][] multiply(int[][] A, int[][] B) { 3 if (A == null || A[0] == null || B == null || B[0] == null) return null; 4 int row = A.length, colA = A[0].length, colB = B[0].length; 5 int[][] C = new int[row][colB]; 6 7 //Map<row, Map<col, val>> 8 Map<Integer, Map<Integer, Integer>> tableB = new HashMap<>(); 9 10 // For each row in matrix B, if there is a non-zero value, put it in the hashMap. 11 for (int k = 0; k < colA; k++) { 12 tableB.put(k, new HashMap<Integer, Integer>()); 13 for (int j = 0; j < colB; j++) { 14 if (B[k][j] != 0) { 15 tableB.get(k).put(j, B[k][j]); 16 } 17 } 18 } 19 20 for (int i = 0; i < row; i++) { 21 for (int k = 0; k < colA; k++) { 22 if (A[i][k] != 0) { 23 for (Integer j : tableB.get(k).keySet()) { 24 C[i][j] += A[i][k] * tableB.get(k).get(j); 25 } 26 } 27 } 28 } 29 return C; 30 } 31 }
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原文地址：http://www.cnblogs.com/beiyeqingteng/p/5735220.html
