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Data Structure lec#2

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Data structure is a very important subject in field of computer. And a very basic subject for the CS and IT student and for electrical engineers TOO.

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Data Structure lec#2

  1. 1. Lecture No.02 Data Structures Zaigham Abbas University of Gujrat (Pakistan)
  2. 2. Implementing Lists  We have designed the interface for the List; we now must consider how to implement that interface.
  3. 3. Implementing Lists  We have designed the interface for the List; we now must consider how to implement that interface.  Implementing Lists using an array: for example, the list of integers (2, 6, 8, 7, 1) could be represented as: A 6 8 7 1 1 2 3 4 5 2 current 3 size 5
  4. 4. List Implementation  add(9); current position is 3. The new list would thus be: (2, 6, 8, 9, 7, 1)  We will need to shift everything to the right of 8 one place to the right to make place for the new element ‘9’. current 3 size 5 step 1: A 6 8 7 1 1 2 3 4 5 2 6 current 4 size 6 step 2: A 6 8 7 1 1 2 3 4 5 2 6 9 notice: current points to new element
  5. 5. Implementing Lists  next(): current 4 size 6 A 6 8 7 1 1 2 3 4 5 2 6 9 5
  6. 6. Implementing Lists  There are special cases for positioning the current pointer: a. past the last array cell b. before the first cell
  7. 7. Implementing Lists  There are special cases for positioning the current pointer: a. past the last array cell b. before the first cell  We will have to worry about these when we write the actual code.
  8. 8. Implementing Lists  remove(): removes the element at the current index current 5 size 6 A 6 8 1 1 2 3 4 5 2 6 9 5 Step 1: current 5 size 5 A 6 8 1 1 2 3 4 5 2 9Step 2:
  9. 9. Implementing Lists  remove(): removes the element at the current index  We fill the blank spot left by the removal of 7 by shifting the values to the right of position 5 over to the left one space. current 5 size 5 A 6 8 1 1 2 3 4 5 2 9Step 2: current 5 size 6 A 6 8 1 1 2 3 4 5 2 6 9 5 Step 1:
  10. 10. Implementing Lists find(X): traverse the array until X is located. int find(int X) { int j; for(j=1; j < size+1; j++ ) if( A[j] == X ) break; if( j < size+1 ) { // found X current = j; // current points to where X found return 1; // 1 for true } return 0; // 0 (false) indicates not found }
  11. 11. Implementing Lists  Other operations: get()  return A[current]; update(X)  A[current] = X; length()  return size; back()  current--; start()  current = 1; end()  current = size;
  12. 12. Analysis of Array Lists  add  we have to move every element to the right of current to make space for the new element.  Worst-case is when we insert at the beginning; we have to move every element right one place.  Average-case: on average we may have to move half of the elements
  13. 13. Analysis of Array Lists  remove  Worst-case: remove at the beginning, must shift all remaining elements to the left.  Average-case: expect to move half of the elements.  find  Worst-case: may have to search the entire array  Average-case: search at most half the array.  Other operations are one-step.
  14. 14. List Using Linked Memory  Various cells of memory are not allocated consecutively in memory.
  15. 15. List Using Linked Memory  Various cells of memory are not allocated consecutively in memory.  Not enough to store the elements of the list.
  16. 16. List Using Linked Memory  Various cells of memory are not allocated consecutively in memory.  Not enough to store the elements of the list.  With arrays, the second element was right next to the first element.
  17. 17. List Using Linked Memory  Various cells of memory are not allocated consecutively in memory.  Not enough to store the elements of the list.  With arrays, the second element was right next to the first element.  Now the first element must explicitly tell us where to look for the second element.
  18. 18. List Using Linked Memory  Various cells of memory are not allocated consecutively in memory.  Not enough to store the elements of the list.  With arrays, the second element was right next to the first element.  Now the first element must explicitly tell us where to look for the second element.  Do this by holding the memory address of the second element
  19. 19. Linked List  Create a structure called a Node. object next  The object field will hold the actual list element.  The next field in the structure will hold the starting location of the next node.  Chain the nodes together to form a linked list.
  20. 20. Linked List  Picture of our list (2, 6, 7, 8, 1) stored as a linked list: 2 6 8 7 1 head current size=5
  21. 21. Linked List Note some features of the list:  Need a head to point to the first node of the list. Otherwise we won’t know where the start of the list is.
  22. 22. Linked List Note some features of the list:  Need a head to point to the first node of the list. Otherwise we won’t know where the start of the list is.  The current here is a pointer, not an index.
  23. 23. Linked List Note some features of the list:  Need a head to point to the first node of the list. Otherwise we won’t know where the start of the list is.  The current here is a pointer, not an index.  The next field in the last node points to nothing. We will place the memory address NULL which is guaranteed to be inaccessible.
  24. 24. Linked List  Actual picture in memory: 1051 1052 1055 1059 1060 1061 1062 1063 1064 1056 1057 1058 1053 1054 2 6 8 7 1 1051 1063 1057 1060 0 head 1054 1063current 2 6 8 7 1 head current 1065

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