Answer:
hope this helps. I am also a learner like you. Please cross check my explanation.
Explanation:
#include
#include
using namespace std;
int main()
{
int a[ ] = {0, 0, 0}; //array declared initializing a0=0, a1=0, a3=0
int* p = &a[1]; //pointer p is initialized it will be holding the address of a1 which means when p will be called it will point to whatever is present at the address a1, right now it hold 0.
int* q = &a[0]; //pointer q is initialized it will be holding the address of a0 which means when q will be called it will point to whatever is present at the address a0, right now it hold 0.
q=p; // now q is also pointing towards what p is pointing both holds the same address that is &a[1]
*q=1
; //&a[0] gets overwritten and now pointer q has integer 1......i am not sure abut this one
p = a; //p is now holding address of complete array a
*p=1; // a gets overwritten and now pointer q has integer 1......i am not sure abut this one
int*& r = p; //not sure
int** s = &q; s is a double pointer means it has more capacity of storage than single pointer and is now holding address of q
r = *s + 1; //not sure
s= &r; //explained above
**s = 1; //explained above
return 0;
}
There are four main learning styles used in classrooms today: Visual, auditory, reading/writing, and kinesthetic.
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.
