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题目：
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 |
链接： http://leetcode.com/problems/sparse-matrix-multiplication/
题解：
Sparse Matrix相乘。题目提示要用HashMap，于是我们就用HashMap， 保存A中不为0行，以及B中不为0的列，然后遍历两个hashmap来更新结果数组。
Time Complexity - O(mnkl)， Space Complexity - O(mn + kl)。
public class Solution { public int[][] multiply(int[][] A, int[][] B) { if(A == null || B == null || A.length == 0 || B.length == 0 || (A[0].length != B.length)) { return new int[][]{}; } Map<Integer, int[]> rowInA = new HashMap<>(); // store non-zero rows in A Map<Integer, int[]> colInB = new HashMap<>(); // store non-zero cols in B for(int i = 0; i < A.length; i++) { for(int j = 0; j < A[0].length; j++) { if(A[i][j] != 0) { rowInA.put(i, A[i]); break; } } } for(int j = 0; j < B[0].length; j++) { for(int i = 0; i < B.length; i++) { if(B[i][j] != 0) { int[] tmp = new int[B.length]; for(int k = 0; k < B.length; k++) { tmp[k] = B[k][j]; } colInB.put(j, tmp); break; } } } int[][] res = new int[A.length][B[0].length]; for(int i : rowInA.keySet()) { for(int j : colInB.keySet()) { for(int k = 0; k < A[0].length; k++) { res[i][j] += rowInA.get(i)[k] * colInB.get(j)[k]; } } } return res; } }
Reference:
311. Sparse Matrix Multiplication
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原文地址：http://www.cnblogs.com/yrbbest/p/5060667.html
