Question

所以我想在数组中输出某个值的位置。如何输出数组值的位置？（代码后的例子）：

public class Journal5b {

public int[] [] createArray (int rSize , int cSize)
{
    int [] [] array = new int [rSize][cSize];
    Random r = new Random();
    for (int[] array1 : array) {
        for (int column = 0; column < array[0].length; column++) {
            array1[column] = r.nextInt(26);
        }
    }

    return array;
}

public void print2DArray (int [] [] array)
{
    for (int[] array1 : array) {
        for (int column = 0; column < array[0].length; column++) {
            System.out.print(array1[column] + "\t");
        }
        System.out.println("\n");
    }
}

public int countInstance (int [] [] array, int searchForNum)
{
    int count = 0;

    for (int row = 0; row < array.length; row++) 
    {
        for (int column = 0; column < array[0].length; column++) {
            if (array[row][column] == searchForNum) {

                count++;
            }
        }
    }
     return count; 
}

public static void main(String[] args) 
{
 Scanner in = new Scanner (System.in);
    int [] [] myArray;
    Journal5b j5b = new Journal5b();
    myArray = j5b.createArray(10, 10);
    j5b.print2DArray(myArray);

   int val;
   System.out.println("Enter the number to search for ");
    val = in.nextInt();
    System.out.println("Your Number popped up " + j5b.countInstance(myArray, val) + " times"); 
} 
}

所以输出我看起来像这样：

22  20  12  25  10  8   3   4   12  25  

11  24  0   7   3   21  23  9   23  11  

22  18  10  24  11  13  0   25  8   18  

 7  14  0   5   11  12  9   22  0   16  

14  11  12  4   16  3   20  22  18  25  

 8  24  12  6   25  4   3   16  10  23  

 23 11  8   12  19  15  3   25  12  6   

 10 3   5   22  11  7   0   7   4   4   

 18 1   14  23  7   13  9   9   12  9   

 20 10  6   14  13  1   9   15  0   3   

 Enter the number to search for 
 10
 Your number popped up 5 times and is located on row 0,2,5,7, 9

如果有意义，我如何输出我想输出的内容...我正在尝试添加到countInstance并找到每行输入的特定数字的位置。

我已经完成了大部分代码，但是我已经使用了这部分代码。我无法找到下一步。我想添加is located on row 0,2,5,7, 9

Answer 1

是的，我现在明白了，对不起，我不得不完全重构你的代码。

public class Journal5b {

   private int[] rowSpotted;//variable to keep rows of searched number
   private int[][] arrayToBeSearched; 

   public class Journal5b (int rows, int columns) {

       rowSpotted = new int[rows];
       arrayToBeSearched = new int[rows][columns];
       this.createArray();
       this.print2dArray();
   }

   private void createArray() {
      Random r = new Random();
      for (int[] array1 : arrayToBeSearched) {
         for (int number: array1) {
               number = r.nextInt(26);
         }
      }
   }

   //I removed the array local variable array so that you won't have to keep passing it as an argument, It's now an instance variable.
   //You can also use a for each loop to populate the inner array, no need to use the for loop.

  public void print2DArray () {
     for (int[] array1 : arrayToBeSearched) {
         for (int num: array1) {
           System.out.print(num + "\t");
         }
      System.out.println("\n");
     }
  }

  //just prints the array to be searched.

  public int countInstance (int searchForNum) {
      int count = 0;

      for (int row = 0; row < arrayToBeSearched.length; row++) {
          for (int column = 0; column < arrayToBeSearched[0].length; column++) {
               if (arrayToBeSearched[row][column] == searchForNum) {
                  rowSpotted[count] = row;  //if the number is found put the row number into the row
                  count++;
               }
          }
      }
      return count; 
  }

 public void printRowSpotted() {
     for(int num : rowSpotted ) {

        System.out.print(num + ',');
     }
 }

 public static void main(String[] args) {
    //scanner stuff......
    Journal5b jb = new Journal5b(10,10);
    //get number to be searched for....
   System.out.print("Your number popped up " + jb.countInstance() + ", on rows " + jb.printRowSpotted());

}

Answer 2

您可以使用#include <list> #include <deque> #include <iostream> using namespace std; template<typename T> ostream & print(const T & start, const T & end) { T tmp = start; for(; tmp != end; ++tmp) { cout<< *tmp<< " "; } return cout; } class A { public: int a; public: A(int a):a(a) {} A(const A & a) {} }; ostream & operator<<(ostream & c, const A & o) { c<<o.a; return c; } int main() { int tab[]={1, 2, 3, 4, 5, 6, 7, 8, 9, 10}; list<A> l1(tab, tab+10); deque<A> d1; list<A>::iterator it; for(it = l1.begin(); it != l1.end(); ++it) { d1.insert(d1.begin(), it[0]); } print(d1.begin(), d1.end())<<endl; return 0; }从ArrayList返回值，而不是使用全局变量。

countInstance

并稍微修改public ArrayList<Integer> countInstance(int[][] array, int searchForNum) { //int count = 0; //Use an ArrayList to get the row index ArrayList<Integer> countList = new ArrayList<>(); for (int row = 0; row < array.length; row++) { for (int column = 0; column < array[0].length; column++) { if (array[row][column] == searchForNum) { //count++; //add row index to the ArrayList count countList.add(row); } } } return countList; }方法以打印所需的结果

main()

数组：输入的某些值的位置

2 个答案: