Ze-Yuan Hu's page

PrintLots

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 operation PrintLots(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 procedure PrintLots(L,P). You should use the basic list operations. What is the running time of your procedure?

Solution

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 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.

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