不用十字鏈表也可以稀疏矩陣相加時間復雜度最壞情況達到O(tuA + tuB);思路比較簡單就不贅述了，代碼如下：

三元組：

package 行邏輯鏈接的順序表實現稀疏矩陣的相乘;

public class Triple {

int row,col;

T v;

public Triple(){}

public Triple(int row,int col, T v){

this.row = row;

this.col = col;

this.v = v;

}

}

構建矩陣存儲結構：

package 行邏輯鏈接的順序表實現稀疏矩陣的相乘;

public class Mat {

final int MAXSIZE = 10;

int mu,nu,tu;

int rpos[] = new int[MAXSIZE + 1];//各行第一個非零元的位置表

Triple data[] = new Triple[MAXSIZE + 1];//Java不支持泛型數組

public Mat(int mu,int nu,int tu){

this.mu = mu;

this.nu = nu;

this.tu = tu;

for(int i=1; i<=MAXSIZE; i++)

data[i] = new Triple();

}

//三元組矩陣的輸出

public void display(){

int i,j,k,m,n,count = 0;

for(i=1; i<=mu; i++){

for(j=1; j<=nu; j++){

for(k=1; k<=tu; k++){

if(i==data[k].row && j==data[k].col){

System.out.print(data[k].v + " ");

count = -1;

break;

}

}

if(count != -1)

System.out.print("0 ");

count = 0;

}

System.out.println();

}

}

}

相加：

package 行邏輯鏈接的順序表實現稀疏矩陣的相乘;

import java.util.*;

public class AddMat {

public static void main(String[] args) {

/*

* @author 王旭

* @time 2014/10/14/23:50

*

*/

Scanner scan = new Scanner(System.in);

int muA,nuA,tuA,muB,nuB,tuB;

System.out.println("請輸入A矩陣的行，列，非零元的個數：");

muA = scan.nextInt();

nuA = scan.nextInt();

tuA = scan.nextInt();

Mat A = new Mat(muA,nuA,tuA);

System.out.println("請輸入A矩陣的三元組：");

for(int i=1; i<=tuA; i++){

A.data[i].row = scan.nextInt();

A.data[i].col = scan.nextInt();

A.data[i].v = scan.nextInt();

}

System.out.println("A矩陣為：");

A.display();

System.out.println("請輸入B矩陣的行，列，非零元的個數：");

muB = scan.nextInt();

nuB = scan.nextInt();

tuB = scan.nextInt();

Mat B = new Mat(muB,nuB,tuB);

System.out.println("請輸入B矩陣的三元組：");

for(int i=1; i<=tuB; i++){

B.data[i].row = scan.nextInt();

B.data[i].col = scan.nextInt();

B.data[i].v = scan.nextInt();

}

System.out.println("B矩陣為：");

B.display();

Mat C = new Mat(muA,nuA,1);

add(A,B,C);

System.out.println("相加后的矩陣C為：");

C.display();

}

public static void add(Mat A, Mat B, Mat C){

int k,l,temp;

//C = new Mat(A.mu,A.nu,1);

k = 1;

l = 1;

while(k

if((A.data[k].row == B.data[l].row) && (A.data[k].col == B.data[l].col)){

temp = A.data[k].v + B.data[l].v;

if(temp != 0){

C.data[C.tu].row = A.data[k].row;

C.data[C.tu].col = A.data[k].col;

C.data[C.tu].v = temp;

C.tu++;

}

k++;

l++;

}

if(( (A.data[k].row == B.data[l].row) && (A.data[k].col < B.data[l].col) ) || (A.data[k].row

C.data[C.tu].row = A.data[k].row;

C.data[C.tu].col = A.data[k].col;

C.data[C.tu].v = A.data[k].v;

C.tu++;

k++;

}

if(( (B.data[l].row == A.data[k].row) && (B.data[l].col < A.data[k].col) ) || (B.data[l].row

C.data[C.tu].row = B.data[l].row;

C.data[C.tu].col = B.data[l].col;

C.data[C.tu].v = B.data[l].v;

C.tu++;

l++;

}

}

}

}

總結

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