Question

我想将字符串矩阵的对角线检索到单独的列表中。请帮我。我整天都在上面。但没有成功。

二维数组结构：

Matrix{{"Toyata","Fortuner","Price", "tuesday", "deliver", "Working"},
        { "Ford","Classic ","Price", "Wedday", "Not Deliver", "Working"}, 
         {"Tata","Jaguar","Price", "Satday", "deliver", "Working" }
        {"Farari","etc","Price", "sunday", "deliver", "NotWorking" }

我想从矩阵上方检索第一行和第三行。即

{"Toyata","Fortuner","Price", "tuesday", "deliver", "Working"},
   {"Tata","Jaguar","Price", "Satday", "deliver", "Working" }

这是我的代码：

    public  Object[][] geCodes(){
   Object [][] Matrix{
               {"Toyata","Fortuner","Price", "tuesday", "deliver", "Working"},
                { "Ford","Classic ","Price", "Wedday", "Not Deliver", "Working"}, 
                 {"Tata","Jaguar","Price", "Satday", "deliver", "Working" }
                {"Farari","etc","Price", "sunday", "deliver", "NotWorking" }};
     Object  Diagonal1=null;
     List<Object> che= new LinkedList<Object>();
        for (int i = 0, j =0; i< matrix.length && j < matrix[0].length; i++, j++) {
             Diagonal1=   matrix[i][j];
             che.add(Diagonal1);

     }
        System.out.println("*******"+che);
                return matrix;

    }

Answer 1

您没有说自己的想法。对角表示第一行的第一个元素，第二行的第二个元素……最后一行的最后一个元素。

您进行一个循环并获得矩阵[i] [i]的位置。您不需要“ i”和“ j”，因为它们是相同的。

如果要获得第一行和第二行，那么就这样做。

Object[][] matrix2 = new Object[2][6];
matrix2[0] = Matrix[0];
matrix2[1] = Matrix[2];

您的代码：

public  Object[][] geCodes(){
    Object [][] Matrix{
        {"Toyata","Fortuner","Price", "tuesday", "deliver", "Working"},
        {"Ford","Classic ","Price", "Wedday", "Not Deliver", "Working"}, 
        {"Tata","Jaguar","Price", "Satday", "deliver", "Working" }
        {"Farari","etc","Price", "sunday", "deliver", "NotWorking" }};


    Object[][] Diagonal1= new Object[2][6];
    Diagonal1[0] = Matrix[0];
    Diagonal1[1] = Matrix[2];

    return matrix; // I don't know why you are returning matrix, 
                   // you probably should be returning the new array, Diagonal1

}

您确实应该使用字符串而不是对象。

Answer 2

如果您想要对角线还是第一个和第三个对角线，我真的不明白，但这是一个示例：

    String[][] matrixStart = {
                {"Toyata", "Fortuner", "Price", "tuesday", "deliver", "Working"},
                {"Ford", "Classic", "Price", "Wedday", "Not Deliver", "Working"},
                {"Tata", "Jaguar", "Price", "Satday", "deliver", "Working"},
                {"Farari", "etc", "Price", "sunday", "deliver", "NotWorking"}};
    List<String[]> matrixEnd = new LinkedList<>();
    for(int i = 0; i < matrixStart.length; i++)
        {
            if(i == 1 || i == 3)
            {
            matrixEnd.add(matrixStart[i]);
            }
        }

    // Printing what's in your matrixEnd
    System.out.println("MatrixEnd :");
    String sysout;
    for(String[] line: matrixEnd)
    {
        sysout = "";
        for(String element: line)
        {
            sysout += element + ", ";
        }
        System.out.println("[" + sysout.substring(0, sysout.length() - 2) + "]");
    }

最后，您会得到类似：

MatrixEnd :
    [Ford, Classic, Price, Wedday, Not Deliver, Working]
    [Farari, etc, Price, sunday, deliver, NotWorking]

此外，您应该使用String而不是Objects并用小写字母命名变量，最后打印“ che”将不会打印内部内容，而只会打印引用，除非您为LinkedList覆盖toString（）。

如何在单独的列表中检索String矩阵的对角线？

2 个答案: