Card 0 of 1
Consider the following code:
int\[\] vals = {6,1,41,5,1};
int\[\]\[\] newVals = new int\[vals.length\]\[\];
for(int i = 0; i < vals.length; i++) {
newVals\[i\] = new int\[vals\[i\]\];
for(int j = 0; j < vals\[i\];j++) {
newVals\[i\]\[j\] = vals\[i\] * (j+1);
}
}
What does the code above do?
Let's look at the main loop in this program:
for(int i = 0; i < vals.length; i++) {
newVals\[i\] = new int\[vals\[i\]\];
for(int j = 0; j < vals\[i\];j++) {
newVals\[i\]\[j\] = vals\[i\] * (j+1);
}
}
The first loop clearly runs through the vals array for the number of items in that array. After this, it creates in the newVals 2D array a second dimension that is as long as the value in the input array vals. Thus, for 41, it creates a 3rd row in newVals that is 41 columns wide.
Then, we go to the second loop. This one iterates from 0 to the value stored in the current location in vals. It places in the given 2D array location the multiple of the value. Think of j + 1. This will go from 1 to 41 for the case of 41 (and likewise for all the values). Thus, you create a 2D array with all the multiples of the initial vals array.
Compare your answer with the correct one above