打印数组中的数字

Question

Im trying to write a function to display all combinations in a jagged array, where each combination contains one element from each sub-array. The jagged array can consist of any number of arrays and each array can have any number of elements. Eg for the following array: a[0] = {1, 3, 5} a[1] = {2, 4} it should return: (1, 2) (1, 4) (3, 2) (3, 4) (5, 2) (5, 4)

I thought of doing it this way but immediately run into trouble. Logically it looks OK to get 1, 2 and 1, 4 but then next run I is set back to 0 (sorry not at devel machine to test now). Can anyone suggest a better solution please?

Here is my code

for (int i = 0; i < array1.length(); i++)
    for (int j = 0; j < array2.length(); j++)

        if (j < array2.length())
            i = 0;
        else 
            i++;

        System.out.println(array1[i] "," array2[j])

Answer 1

You don't need this:

if (j < array2.length())
            i = 0;
        else 
            i++;

i is incremented automatically in a for loop.

This should be fine:

for (int i = 0; i < array1.length(); i++)
    for (int j = 0; j < array2.length(); j++)
        System.out.println(array1[i] "," array2[j])

Answer 2

If I'm understanding your question correctly (which I might not be) I think all you need is just

for (int i = 0; i < array1.length(); i++){
  for (int j = 0; j < array2.length(); j++){
    System.out.println(array1[i] "," array2[j]);
  }
}

to achieve the desired result

Answer 3

How about this:

int a [] = {1,2,3}; int b[] = {1,2};

for (int i = 0; i < b.length; i++) {
    for (int j = 0; j < a.length; j++) {
        System.out.println(a[i]+","+a[j]);

    }

}

Answer 4

Your if statement inside the loop breaks everything. You just need 2 nested loop to complete your task:

for (int i = 0; i < array1.length(); i++)
    for (int j = 0; j < array2.length(); j++) {
        System.out.println(array1[i] + "," + array2[j]);
    }
}

Answer 5

for (int i = 0; i < array1.length(); i++)
    for (int j = 0; j < array2.length(); j++)
        System.out.println("(" + array1[i] + "," array2[j] + ")");

Answer 6

Here's a general solution that works with any number of arrays (beware the exponential nature of the runtime of this algorithm):

int[][] arrays = new int[][]
{
    {1, 2, 3, 4, 5, 6},
    {1, 2, 3, 4, 5, 6}, 
    {1, 2, 3, 4, 5, 6}
}; // let's print all fair rolls of a 3d6

if (arrays.length == 0) return; // this is why we can't have nice things

int[] currentPos = new int[arrays.length];

while(currentPos[arrays.length - 1] != arrays[arrays.length - 1].length)
{
    // print the current value
    System.out.print(arrays[0][currentPos[0]]);
    for (int i = 1; i < arrays.length; ++i)
        System.out.print(", " + arrays[i][currentPos[i]]);
    System.out.println();

    // increment the "counter"
    ++currentPos[0];
    for (int i = 1; i < arrays.length; ++i)
    {
        if (currentPos[i - 1] == arrays[i - 1].length)
        {
            currentPos[i - 1] = 0;
            ++currentPos[i];
        }
        else break;
    }
}