kruti-working-branch #14
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Good job <3 !!
@ -37,1 +31,3 @@
return element;
E currentValue = start.data;
start = start.next;
return currentValue;
Case where only 1 element in queue, end pointer is dangling.
@ -15,3 +17,3 @@
}
// Will return true if size is 0
Node<E> head = null;
Why do you need two pointers, both insertion and deletion happen at head
@ -23,2 +19,2 @@
mergesort(arr, aux, mid+1, hi);
merge(arr, aux, lo, mid, hi);
// Copy array
E[] auxArr = (E[]) new Object[high_index - low_index + 1];
Making a new aux arry with each recursive call is expensive
@ -94,0 +51,4 @@
queueArray = (E[]) new Object[targetSize];
int j = 0;
for(int i=startPtr; i<endPtr; i++)
queueArray[j++]=tempArray[i];
If the old queue is wrapped, then endPtr will be less than startPtr
@ -103,0 +74,4 @@
@Override
public E next() {
return queueArray[iteratorIndex++];
Iterator does not account for wrapping
Pull request closed