int findMax(int *array)
{
int max = array[0];
for(int i = 1; i < array.length(); i++)
{
if(array[i] > max)
max = array[i]
}
return max;
}
If you are using an array : sort using qsort() then take middle element.
maxValue = function (array) {mxm = array[0];for (i=0; i<array.length; i++) {if (array[i]>mxm) {mxm = array[i];}}return mxm;}; i don't know
To find the highest value in an array, start with the first element as the initial maximum. Iterate through each element in the array, comparing it to the current maximum. If an element is greater than the current maximum, update the maximum to this element. After checking all elements, the current maximum will be the highest value in the array.
Usually one element at a time. If you want to process all elements of an array, you write a loop.Usually one element at a time. If you want to process all elements of an array, you write a loop.Usually one element at a time. If you want to process all elements of an array, you write a loop.Usually one element at a time. If you want to process all elements of an array, you write a loop.
The simplest way is probably to read the numbers into an array and then prints each element of the array starting at the last one and moving backwards.
If you are using an array : sort using qsort() then take middle element.
maxValue = function (array) {mxm = array[0];for (i=0; i<array.length; i++) {if (array[i]>mxm) {mxm = array[i];}}return mxm;}; i don't know
To find the highest value in an array, start with the first element as the initial maximum. Iterate through each element in the array, comparing it to the current maximum. If an element is greater than the current maximum, update the maximum to this element. After checking all elements, the current maximum will be the highest value in the array.
Usually one element at a time. If you want to process all elements of an array, you write a loop.Usually one element at a time. If you want to process all elements of an array, you write a loop.Usually one element at a time. If you want to process all elements of an array, you write a loop.Usually one element at a time. If you want to process all elements of an array, you write a loop.
The simplest way is probably to read the numbers into an array and then prints each element of the array starting at the last one and moving backwards.
The maximum number of comparisons required in a binary search algorithm to find a specific element in a sorted array is log(n), where n is the number of elements in the array.
/* using ellipses (...) to indicate tabs for clarity */ double largest (double *array, int M, int N) { ... int i, j; ... double *element; ... double answer = array[0][0]; ... for (i=0; i<M; i++) { ... ... for (j=0; j<N; j++) { ... ... ... element = array + i*M + j; ... ... ... if (*element > answer) answer = *element; ... ... } ... } ... return answer; }
== Java does not allow reference to memory locations. == In C: for (i=0; i<n; ++i) printf ("a[%d] is at %p\n", i, &a[i]);
The results of that programming error is undefined. You must NEVER EVER write, or EVEN READ an array element beyond the allocated size of the array. Period.I would flunk a student that consistently did this, and I would fire a programmer that did the same.
If you have the series stored in an array, you loop through the array and print each array element in turn. Another possibility is to print out the numbers in the series as you generate them. In that case, you may not need to store anything (depending on the series, of course).
int max(int arr[], int arrSize){int maximum = arr[0];for (int i = 0; i < arrSize; i++){if (maximum < arr[i]){maximum = arr;}}return maximum;}
// Assuming you dynamically allocated this array using "new"... delete array[arraysize - 1]; arraysize--;