Brandi Love

Ranch Hand

Posts: 133

posted 13 years ago

Your i loop iterates over each element of the 1d array. The j and k loops visit every element of the 2d array in turn and assign it into the i'th element of the 1d array. The very last assignment is to assign the last 2d element (3) to the i'th element of the 1d array, so every element ends up being 3 -- plus you've done 20 times more assignments than are necessary!

To fix: remove the loop over i altogether. Inside the k loop, compute the value of i (i.e., the index into the 1d array) as a function of j and k. The right equation is

i = j * array[0].length + k

To fix: remove the loop over i altogether. Inside the k loop, compute the value of i (i.e., the index into the 1d array) as a function of j and k. The right equation is

i = j * array[0].length + k

posted 13 years ago

The problem is that you're looping on i. What you're told the computer to do is:

For each 1 from 1 to 20, place each element of array[j][k] into sort[i].

Therefore, each time through the outermost loop, you place each value in sort[i]. The final element is a 3, so that is what gets left there.

Consider not using an outer foor loop and instead assigning i = 0 to start. Then, after you populate sort[i], increment i by one.

For each 1 from 1 to 20, place each element of array[j][k] into sort[i].

Therefore, each time through the outermost loop, you place each value in sort[i]. The final element is a 3, so that is what gets left there.

Consider not using an outer foor loop and instead assigning i = 0 to start. Then, after you populate sort[i], increment i by one.

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