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