[LeedCode]Sparse Matrix Multiplication

这个题目考点在于矩阵中有很多0的时候怎么处理。我是创建了一个Node的数据结构，分别存index和值。看网上有个小哥是把index和值分别存在 i 和i + 1 的位置，感觉也很好，不过算法复杂度本质没有变化。

public class Solution {
    public int[][] multiply(int[][] A, int[][] B) {
        int row_A = A.length;
        int col_A = A[0].length;
        int row_B = B.length;
        int col_B = B[0].length;
        int[][] result = new int[row_A][col_B];
        for (int r = 0; r < row_A; r++) {
            List<Node> list = new ArrayList<Node>();
            for (int c = 0; c < col_A; c++) {
                if (A[r][c] != 0) {
                    list.add(new Node(c, A[r][c]));
                }
            }
            for (int c = 0; c < col_B; c++) {
                result[r][c] = helper(list, B, c);
            }
        }
        return result;
    }
    public int helper(List<Node> A, int[][] B, int c) {
        int result = 0;
        for (Node node : A) {
            result += node.val * B[node.ind][c];
        }
        return result;
    }
    class Node {
        int ind;
        int val;
        public Node(int ind, int val) {
            this.ind = ind;
            this.val = val;
        }
    }
}

后来还是写了一下，因为发现我的要300ms。这样来说同样的复杂度，多调用一个类和一个helper函数还是会很影响速度的。代码如下，大概只需要100ms

public class Solution {
    public int[][] multiply(int[][] A, int[][] B) {
        int row_A = A.length;
        int col_A = A[0].length;
        int row_B = B.length;
        int col_B = B[0].length;
        int[][] result = new int[row_A][col_B];
        for (int r = 0; r < row_A; r++) {
            List<Integer> list = new ArrayList<Integer>();
            for (int c = 0; c < col_A; c++) {
                if (A[r][c] != 0) {
                    list.add(c);
                    list.add(A[r][c]);
                }
            }
            for (int c = 0; c < col_B; c++) {
                int tmp = 0;
                for (int i = 0; i < list.size(); i += 2) {
                    tmp += list.get(i + 1) * B[list.get(i)][c];
                }
                result[r][c] =tmp;
            }
        }
        return result;
    }
}