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311. Sparse Matrix Multiplicatio

311. Sparse Matrix Multiplicatio

作者: Jeanz | 来源:发表于2017-08-21 08:57 被阅读0次

    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 |
    

    一刷
    题解:已经强调了是sparse matrices, 于是要考虑用其他的方法

    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.

    例如A的第0行第0列的value为1, 那么A[0] = {0, 1};
    A的第1行第0列的value为-1,第2列的value为3, 那么A[1] = {0, -1, 2, 3};

    然后把A中的一个个数字取出来,求解。

    class Solution {
        public int[][] multiply(int[][] A, int[][] B) {
        int m = A.length, n = A[0].length, nB = B[0].length;
        int[][] result = new int[m][nB];
    
        List[] indexA = new List[m];
        for(int i = 0; i < m; i++) {
            List<Integer> numsA = new ArrayList<>();
            for(int j = 0; j < n; j++) {
                if(A[i][j] != 0){
                    numsA.add(j); 
                    numsA.add(A[i][j]);
                }
            }
            indexA[i] = numsA;
        }
    
        for(int i = 0; i < m; i++) {
            List<Integer> numsA = indexA[i];
            for(int p = 0; p < numsA.size() - 1; p += 2) {
                int colA = numsA.get(p);
                int valA = numsA.get(p + 1);
                for(int j = 0; j < nB; j ++) {
                    int valB = B[colA][j];
                    result[i][j] += valA * valB;
                }
            }
        }
    
        return result;   
    }
    }
    

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