GaussianElimination 高斯消去法java（求逆，求解）

java代码

import java.util.Scanner;
public class GaussianElimination{

    public static void Input(int n, double[][] A, double[] b, int length){
        Scanner in = new Scanner(System.in);
        System.out.println("输入方程组的系数矩阵A;");
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                A[i][j] = in.nextDouble();
            }
        }
        if(length == 1){
            System.out.println("输入方程组矩阵b：");
            for (int i = 0; i < n; i++) {
                b[i] = in.nextDouble();
            }
        }
    }
    public static void Merge1(int n, double[][] A, double[] b){
        for (int i = 0; i < n; i++) {
            A[i][n] = b[i];
        }
        Print(n, A, 1);
        return;
    }
    public static void Merge2(int n, double[][] A){
        for (int i = 0; i < n; i++) {
            for (int j = n; j < 2*n; j++) {
                A[i][j] = 0;
            }
             A[i][n+i] = 1;
        }
        Print(n, A, n);
        return;
    };
    public static void Print(int n, double[][] A, int length){
        System.out.println("##################手 动 分 割 线#####################");
        System.out.println("增广矩阵：");
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n + length; j++) {
                System.out.printf("%-11.4f", A[i][j]);
            }
            System.out.print("\n");
        }
    } //length为1，求逆为n
    public static void Swap(int n, double[][]A, int i, int pivotrow, int length){
        double temp = 0;
        for (int k = 0; k < n+length; k++) {
            temp = A[i][k];
            A[i][k] = A[pivotrow][k];
            A[pivotrow][k] = temp;
        }
    } //交换行
    public static int SelectPivotrow(int n, double[][]A, int i) {
        int pivotrow = i;
        for (int j = i + 1; j < n; j++) {
            if (Math.abs(A[j][i]) > Math.abs(A[pivotrow][i]))
                pivotrow = j;
        }
        if (A[pivotrow][i] == 0)
            return -1;
        else
            return pivotrow;
    } //选择主元
    public static void ForwardElimination(int n, double[][]A, int length){
        for (int i = 0; i < n-1; i++) {
            int pivotrow = SelectPivotrow(n, A, i);
            if(pivotrow == -1){
                System.out.println("输出结果不唯一");
                break;
            }
            Swap(n, A, i, pivotrow, length);
            for (int j = i+1; j < n; j++) {
                double temp = A[j][i]/A[i][i];
                for (int k = i; k < n+length; k++) {
                    A[j][k] = A[j][k] - A[i][k] * temp;
                }
            }
            Print(n, A, length);
        }
    } //前向消去，求length为1，求逆为n
    public static void BackElimination(int n, double[][]A, int length){
        for (int i = n-1; i > 0; i--) {
            for (int j = i-1; j >= 0; j--) {
                double temp = A[j][i]/A[i][i];
                for (int k = n+length-1; k > j; k--) {
                    A[j][k] = A[j][k] - A[i][k] * temp;
                }
            }
            Print(n, A, length);
        }
    } //反向替换，求length为1，求逆为n
    public static void Normalization(int n, double[][]A, int length){
        for (int i = 0; i < n; i++) {
            for (int j = n; j < 2*n; j++) {
                A[i][j] = A[i][j]/A[i][i];
            }
            A[i][i] = 1;
        }
        Print(n, A, length);
    }  //归一化,length为1，求逆为n
    public static void PrintResult(int n, double[][] A, int length){
        System.out.println("##################手 动 分 割 线#####################");
        System.out.println("最终结果为：");
        if(length == 1){
            for (int i = 0; i < n; i++) {
                System.out.println("x"+(i+1)+" = "+A[i][n+length-1]);
            }
        }
        else if(n == length){
            System.out.println("矩阵的逆：");
            for (int i = 0; i < n; i++) {
                for (int j = n; j < 2*n; j++) {
                    System.out.printf("%-11.2f", A[i][j]);
                }
                System.out.print("\n");
            }
        }
        return ;
    }
    public static void Do(int n, double[][]A, int length){
        ForwardElimination(n, A, length);
        BackElimination(n, A, length);
        Normalization(n, A, length);
        PrintResult(n, A, length);
    }

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        while(true){
            System.out.println("-----------------高 斯 消 元 法------------------");
            System.out.println("求逆矩阵请输入1，求解请输入2，退出请输入0。");
            int flag = in.nextInt();
            if(flag != 0){
                System.out.println("输入方程组的元数：");
                int n = in.nextInt();
                double[][] A = new double[n][2*n];
                double[] b = new double[n];
                double[] x = new double[n];
                double[][] A_I = new double[n][n];
                if(flag ==1){
                    Input(n, A, b, n);
                    Merge2(n, A);
                    Do(n, A, n);
                }else{
                    Input(n, A, b ,1);
                    Merge1(n, A, b);
                    Do(n, A,1);
                }
            }else break;
        }
        System.out.println("-----------------程 序 结 束 !!------------------");
    }

}