311. Sparse Matrix Multiplication(有锁)【分治|循环】
2017/3/26 16:09:31
问题：求解稀疏矩阵乘法
版本1：朴素算法 θ(n3) 采用常规矩阵乘法公式
public static int[][] SparseMatrixMultiplication( int[][] A , int[][] B ){
int M = A.length;
int N = B[0].length;//取列长
int K = A[0].length;
int[][] C = new int[M][N];
for ( int i=0; i<M;i++ )
for ( int j=0;j<N;j++ ){
C[i][j] = 0;
for ( int k=0;k<K;k++ )
C[i][j] += A[i][k] * B[k][j];
}
return C;
}
版本2：分治法 θ(n3) 利用分块矩阵乘法性质
public static void main(String[] args) {
int[][] A = { {1,2,3},{4,5,6}};
int[][] B = { {1,2},{3,4},{5,6}};
int[][] C = SparseMatrixMultiplication(A, B);
for ( int i=0;i<C.length;i++ ){
for ( int j=0;j<C[0].length;j++ )
System.out.print(C[i][j]+" ");
System.out.println();
}
}
public static class Square{
public Square(int rowStart,int rowEnd,int colStart,int colEnd){
this.rowStart = rowStart;
this.rowEnd = rowEnd;
this.colStart = colStart;
this.colEnd = colEnd;
}
int rowStart;
int rowEnd;
int colStart;
int colEnd;
}
public static int[][] SparseMatrixMultiplication( int[][] A , int[][] B ){
Square a = new Square( 0 , A.length - 1 , 0 , A[0].length - 1);
Square b = new Square( 0 , B.length - 1 , 0 , B[0].length - 1);
return SparseMatrixMultiplicationHelper( A , B , a , b );
}
public static int[][] MatrixSum( int[][] A , int[][] B){
int[][] C = new int[A.length][A[0].length];
for( int i=0;i<C.length;i++ )
for( int j=0;j<C[0].length;j++ )
C[i][j] = A[i][j] + B[i][j];
return C;
}
public static int[][] MatrixMerge( int[][] c11 , int[][] c12 , int[][] c21 ,int[][] c22){
int[][] C = new int[c11.length+c21.length][c11[0].length+c12[0].length];
for ( int i=0;i<c11.length;i++){
for( int j=0;j<c11[0].length;j++ ){
C[i][j] = c11[i][j];
}
}
for ( int i=0;i<c12.length;i++){
for( int j=0;j<c12[0].length;j++ ){
C[i][c11[0].length+j] = c12[i][j];
}
}
for ( int i=0;i<c21.length;i++){
for( int j=0;j<c11[0].length;j++ ){
C[c11.length+i][j] = c21[i][j];
}
}
for ( int i=0;i<c22.length;i++){
for( int j=0;j<c22[0].length;j++ ){
C[c11.length+i][c11[0].length+j] = c22[i][j];
}
}
return C;
}
public static int[][] SparseMatrixMultiplicationHelper( int[][] A , int[][] B ,Square a , Square b){
//递归基
int M = a.rowEnd - a.rowStart + 1 ;
int N = b.colEnd - b.colStart + 1;
int[][] C = new int[M][N];
if ( a.rowStart == a.rowEnd || a.colEnd == a.colStart ||
b.colStart == b.colEnd || b.rowStart == b.rowEnd ) {
if ( a.rowStart == a.rowEnd && b.colStart == b.colEnd ){
for ( int i=0;i<b.rowEnd-b.rowStart+1;i++ ){
C[0][0] += A[a.rowStart][a.colStart+i] * B[b.rowStart+i][b.colStart];
}
}
else if ( a.colStart == a.colEnd && b.rowEnd == b.rowStart ){
for ( int i=0;i<a.rowEnd-a.rowStart+1;i++ ){
for ( int j=0;j<b.colEnd-b.colStart+1;j++){
C[i][j] = A[a.rowStart+i][a.colStart] * B[b.rowStart][b.colStart+j];
}
}
}
return C;
}
Square a11 = new Square(a.rowStart , (a.rowEnd+a.rowStart)/2 , a.colStart , (a.colEnd+a.colStart)/2);
Square a12 = new Square(a.rowStart , (a.rowEnd+a.rowStart)/2 , (a.colEnd+a.colStart)/2+1 , a.colEnd);
Square a21 = new Square((a.rowEnd+a.rowStart)/2+1 , a.rowEnd , a.colStart , (a.colEnd+a.colStart)/2);
Square a22 = new Square((a.rowEnd+a.rowStart)/2+1 , a.rowEnd , (a.colEnd+a.colStart)/2+1 , a.colEnd);
Square b11 = new Square(b.rowStart , (b.rowEnd+b.rowStart)/2 , b.colStart , (b.colEnd+b.colStart)/2);
Square b12 = new Square(b.rowStart , (b.rowEnd+b.rowStart)/2 , (b.colEnd+b.colStart)/2+1 , b.colEnd);
Square b21 = new Square((b.rowEnd+b.rowStart)/2+1 , b.rowEnd , b.colStart , (b.colEnd+b.colStart)/2);
Square b22 = new Square((b.rowEnd+b.rowStart)/2+1 , b.rowEnd , (b.colEnd+b.colStart)/2+1 , b.colEnd);
int[][] c11 = MatrixSum(SparseMatrixMultiplicationHelper(A,B,a11,b11),SparseMatrixMultiplicationHelper(A,B,a12,b21));
int[][] c12 = MatrixSum(SparseMatrixMultiplicationHelper(A,B,a11,b12),SparseMatrixMultiplicationHelper(A,B,a12,b22));
int[][] c21 = MatrixSum(SparseMatrixMultiplicationHelper(A,B,a21,b11),SparseMatrixMultiplicationHelper(A,B,a22,b21));
int[][] c22 = MatrixSum(SparseMatrixMultiplicationHelper(A,B,a21,b12),SparseMatrixMultiplicationHelper(A,B,a22,b22));
C = MatrixMerge( c11, c12 ,c21 , c22);
return C;
}
版本3： 既然是稀疏矩阵，大部分元素是0，所以可以剪掉多余的乘法运算。A的某一行全为0或者B的某一列全为0都可以省略计算。
Python
class Solution(object):
def multiply(self, A, B):
if len(A)==0 or len(B)==0: return 0
m, n = len(A), len(B[0])
res = [ [0]*n for i in range(m) ]
zerom, zeron = [True]*m, [True]*n
for i in range(m):
for index in range(len(A[0])):
if A[i][index]!=0:
zerom[i] = False
break
for i in range(n):
for index in range(len(B)):
if B[index][i]!=0:
zeron[i] = False
break
for i in range(m):
if zerom[i]==True: continue
for j in range(n):
if zeron[j]==True: continue
for mul_ind in range(len(A[0])):
res[i][j] += A[i][mul_ind] * B[mul_ind][j]
return res
