写一段用分治法求解矩阵乘法的java代码
时间: 2023-06-08 17:03:30 浏览: 60
好的,下面是用分治法求解矩阵乘法的Java代码:
```
public class MatrixMultiplication {
public int[][] multiply(int[][] A, int[][] B) {
int m = A.length, n = A[0].length, k = B[0].length;
int[][] res = new int[m][k];
matrixMultiply(A, B, res, 0, 0, 0, 0, m, n, k);
return res;
}
private void matrixMultiply(int[][] A, int[][] B, int[][] C,
int rA, int cA, int rB, int cB,
int m, int n, int k) {
if (n != k) {
throw new IllegalArgumentException("Matrix size mismatch");
}
if (m == 1 && n == 1 && k == 1) {
C[rA][cB] += A[rA][cA] * B[rB][cB];
return;
}
int newM = m / 2, newN = n / 2, newK = k / 2;
matrixMultiply(A, B, C, rA, cA, rB, cB, newM, newN, newK);
matrixMultiply(A, B, C, rA, cA + newN, rB + newN, cB, newM, newN, newK);
matrixMultiply(A, B, C, rA, cA, rB, cB + newK, newM, newN, newK);
matrixMultiply(A, B, C, rA, cA + newN, rB + newN, cB + newK, newM, newN, newK);
matrixMultiply(A, B, C, rA + newM, cA, rB, cB, newM, n - newN, newK);
matrixMultiply(A, B, C, rA + newM, cA + newN, rB + newN, cB, newM, n - newN, newK);
matrixMultiply(A, B, C, rA + newM, cA, rB, cB + newK, newM, n - newN, newK);
matrixMultiply(A, B, C, rA + newM, cA + newN, rB + newN, cB + newK, newM, n - newN, newK);
}
}
```
这个算法的时间复杂度是 $O(n^3)$,但是分治法的优化使得它可以有效地处理大规模矩阵乘法运算。
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