311 Sparse Matrix Multiplication
Given twosparse 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 |

这题的终极版:下面一个算法,只利用了A是sparse的特性。如果要利用B也sparse的特性的话,我们可以把内层循环的数目减少。其实我们不用把j从头到尾loop一遍,我们用O(n2)的时间把不为0的j的位置找出来,内层循环,只loop那些不为0的就ok了。听说,matrix multiplication是O(n2.几)的算法,不能优化为O(n2)的。
/**
* @param A: a sparse matrix
* @param B: a sparse matrix
* @return: the result of A * B
*/
public int[][] multiply(int[][] A, int[][] B) {
if (A == null || B == null) {
return null;
}
int n = A.length;
int m = B[0].length;
int[][] res = new int[n][m];
List<List<Integer>> nonZeroB = new ArrayList<>();
for (int i = 0; i < B.length; i++) {
nonZeroB.add(new ArrayList<>());
for (int j = 0; j < m; j++) {
if (B[i][j] != 0) {
nonZeroB.get(i).add(j);
}
}
}
for (int i = 0; i < A.length; i++) {
for (int k = 0; k < A[0].length; k++) {
if (A[i][k] == 0) {
continue;
}
for (Integer j : nonZeroB.get(k)) {
res[i][j] += A[i][k] * B[k][j];
}
}
}
return res;
}
这题第一种想到的方法十分之慢,就是直接按照矩阵乘法去做。虽然发现了0就直接跳过,但这样效率还是很低。(贴上来做参考)后来看了discuss答案之后发现,应该把check 0的那一个loop抽出来。这样就可以把一行/一列直接跳过。
public int[][] multiply(int[][] A, int[][] B) {
if (A == null && B == null) {
return null;
}
int n = A.length;
int m = B[0].length;
int[][] result = new int[n][m];
for (int i = 0; i < n; i++) {
for (int k = 0; k < A[0].length; k++) { // pull this loop out to check 0 earlier
if (A[i][k] == 0) {
continue;
}
for (int j = 0; j < m; j++) {
if (B[k][j] == 0) { // 其实这一步每次只优化了一句话,所以没什么鸟用
continue;
}
result[i][j] += A[i][k] * B[k][j];
}
}
}
return result;
}
很慢的:
public int[][] multiply(int[][] A, int[][] B) {
if (A == null && B == null) {
return null;
}
int n = A.length;
int m = B[0].length;
int[][] result = new int[n][m];
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
for (int k = 0; k < A[0].length; k++) {
if (A[i][k] == 0 || B[k][j] == 0) {
continue;
}
result[i][j] += A[i][k] * B[k][j];
}
}
}
return result;
}
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