## Problem

Today, I finished the problem 3.2. The question is following:

You are given a linked list, L, and another linked list, P, containing integers sorted in ascending order. The operationPrintLots(L,P)will print the elements in L that are in positions specified by P. For instance, if P = 1,3,4,6, the first, third, fourth, and sixth elements in L are printed. Write the procedurePrintLots(L,P). You should use the basic list operations. What is the running time of your procedure?

## Solution

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 | ```
void
printLots(List L, List P)
{
Pos dummyP, dummyL; // creates dummy nodes to traverse the list
int i = 0, idx, outofelement;
dummyP = P->Next;
dummyL = L->Next;
while (dummyP != NULL)
{
idx = dummyP->Element;
if (idx >= 0)
{
// if the idx is larger or equal to where the dummyL currently is
// we don't want to reset the dummyL to the very beginning of
// the list L again to redo the traverse.
if (idx < i)
{
dummyL = L->Next;
i = 0;
}
for(; i < idx; i++)
{
if (dummyL->Next != NULL)
{
dummyL = dummyL->Next;
}
else
{
outofelement = 1;
break;
}
}
if (outofelement == 1)
{
printf("No element in position %d, ", idx);
}
else{
printf("%d, ", dummyL->Element);
}
}
else
{
exit(EXIT_FAILURE);
}
outofelement = 0;
dummyP = dummyP->Next;
}
}
``` |

The problem isn't hard to solve. However, to get things right, I need to develop several test cases. Let's develop a solution that can handle more general situation. In other words, linked list, P, doesn't necessarily contain integers sorted in ascending order. Here are test cases I developed:

L: 23, 44, 45, 57, 89, -1 P: 1, 3, 4, 5 <--- normal case 1, 3, 4, 6 <--- there is no sixth element in L 1, 3, 4, 6, 7 <--- there is no sixth, seventh element in L 6, 7, 3, 1 <--- there is no sixth, seventh element in L, but have third, first element 6, 2, 7, 1 <--- a no element (6th) followed by a existing element (2nd) -9, 1, 3, 4 <--- negative integer from P appears at the beginning 1, 2, 4, -10 <--- negative integer from P appears at the end

The code presented above handles all these different situations. In addition, if the integers presented in P
are actually in ascending order, we want to take advantage of this piece of information. That's why we check
`if (idx < i)`. We don't want to reset the traverse ptr (i.e. `dummyL`) every single time. In other words,
if the number in P is actually ascending, we want to move the traver ptr from its current pos instead of reset.