Answer:
1) The value of x will be 6.
2) The value of y will be 7.
Explanation:
1) The value of x will be 6.
The enum values are labeled by default from 1. This means that Sun = 1, Mon = 2, Tue = 3 and so on.
So, x = Mon = 2 and y = Sat = 6
x increases up to y = 6 in the while loop.
and then y also increments by 1.
2)So the value of y = 7.
You will need actual printf("Sun") or printf("Mon") for printing the actual text for the enum.
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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.
Answer:
The answer to this question is given below in the explanation section.
Explanation:
The for-loop given in the question is:
for ( j = 0; j < 10; j++ )
{
appendItem (myList, aNumber); //this loop append a number to a list myList
}
This loop starts from J variable's value zero and when J's value is less than 10, the loop iterate through its body until J's value becomes greater or equal to 10. As J's value exceed nine, the loop will get terminated.
So this loop repeats 10 times its loop body, at the 11th time, the condition becomes false and the loop will get terminated.
The answer & explanation for this question is given in the attachment below.
