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 |
70ms solution:
1 public class Solution { 2 public int[][] multiply(int[][] A, int[][] B) { 3 int m = A.length; 4 int n = A[0].length; 5 int bn = B[0].length; 6 int[][] C = new int[m][bn]; 7 8 for (int i=0; i<m; i++) { 9 for (int k=0; k<n; k++) { 10 if (A[i][k] == 0) continue; 11 for (int j=0; j<bn; j++) { 12 C[i][j] += A[i][k] * B[k][j]; 13 } 14 } 15 } 16 return C; 17 } 18 }
use one HashTable: 160 ms
The idea is derived from a CMU lecture.
A sparse matrix can be represented as a sequence of rows, each of which is a sequence of (column-number, value) pairs of the nonzero values in the row.
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 m = A.length, n = A[0].length, l = B[0].length; 5 int[][] C = new int[m][l]; 6 Map<Integer, HashMap<Integer, Integer>> tableB = new HashMap<>(); 7 8 for(int k = 0; k < n; k++) { 9 tableB.put(k, new HashMap<Integer, Integer>()); 10 for(int j = 0; j < l; j++) { 11 if (B[k][j] != 0){ 12 tableB.get(k).put(j, B[k][j]); 13 } 14 } 15 } 16 17 for(int i = 0; i < m; i++) { 18 for(int k = 0; k < n; k++) { 19 if (A[i][k] != 0){ 20 for (Integer j: tableB.get(k).keySet()) { 21 C[i][j] += A[i][k] * tableB.get(k).get(j); 22 } 23 } 24 } 25 } 26 return C; 27 } 28 }
时间: 2024-10-01 22:44:58