311. Sparse Matrix Multiplication

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: