Since both arrays are already sorted, that means that the first int of one of the arrays will be smaller than all the ints that come after it in the same array. We also know that if the first int of arr1 is smaller than the first int of arr2, then by the same logic, the first int of arr1 is smaller than all the ints in arr2 since arr2 is also sorted.
public static int[] merge(int[] arr1, int[] arr2) {
int i = 0; //current index of arr1
int j = 0; //current index of arr2
int[] result = new int[arr1.length+arr2.length]
while(i < arr1.length && j < arr2.length) {
result[i+j] = Math.min(arr1[i], arr2[j]);
if(arr1[i] < arr2[j]) {
i++;
} else {
j++;
}
}
boolean isArr1 = i+1 < arr1.length;
for(int index = isArr1 ? i : j; index < isArr1 ? arr1.length : arr2.length; index++) {
result[i+j+index] = isArr1 ? arr1[index] : arr2[index]
}
return result;
}
So this implementation is kind of confusing, but it's the first way I thought to do it so I ran with it. There is probably an easier way, but that's the beauty of programming.
A quick explanation:
We first loop through the arrays comparing the first elements of each array, adding whichever is the smallest to the result array. Each time we do so, we increment the index value (i or j) for the array that had the smaller number. Now the next time we are comparing the NEXT element in that array to the PREVIOUS element of the other array. We do this until we reach the end of either arr1 or arr2 so that we don't get an out of bounds exception.
The second step in our method is to tack on the remaining integers to the resulting array. We need to do this because when we reach the end of one array, there will still be at least one more integer in the other array. The boolean isArr1 is telling us whether arr1 is the array with leftovers. If so, we loop through the remaining indices of arr1 and add them to the result. Otherwise, we do the same for arr2. All of this is done using ternary operations to determine which array to use, but if we wanted to we could split the code into two for loops using an if statement.
It changes a little depending on what programming language you're using, but in C you could say
int times_ten (int num) {
num = num*10;
return num;
}
Or, in general terms:
Integer Function times_ten (Integer num)
Set num = num * 10
Return num
This is all done assuming the number used as an argument is an integer and that you are returning an integer value, not a decimal. The main thing to notice is that since you have to return a value, you must have two things: a return statement, and a type declaration for the function, otherwise you'll get an error.
If Darren wants to substitute every occurrence of the word bulky in his spreadsheet with the word strong, the one that he should use is the Find all function. This function simplifies the use of the Find method. Therefore, the correct answer would be option B.
Answer:
The answer is "True".
Explanation:
Traditional techniques of application development were focused on pre-development stages. These control system flows through claims to design and upgrades and will then tests and repair are directional there.
- It enables the user to create and run a machine-based data system.
- It uses program and device, that specifically the solutions, that enable manual information to be processed.
Answer:
public class TicTacToe //Defining TicTacToe class
{
char board[3][3] ; //Creating a 2D array instance variable
TicTacToe() //Constructor to initialize the array with " - "
{
for( int i = 0;i<3;i++) //Loop for the row of array
{
for(int j = 0;j<3;j++) //Loop for the column of array
{
Board[i][j] = '-'; //Assigning "-" in each cell
}
}
}
public char getter() //Defining getter() method to return the array
{
return Board; //returning the array
}
}
