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Chapter 3

Stacks. Chapter 3. Linear Lists. Operations are; Insertion Deletion Retrieval Traversal (exception for restristed lists). Figure 3-2. Stacks. A stack is a Last in First out (LIFO) data structure in which all insertions and deletions are restricted to one end called top. Figure 4-1.

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Chapter 3

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  1. Stacks Chapter 3

  2. Linear Lists Operations are; Insertion Deletion Retrieval Traversal (exception for restristed lists). Figure 3-2

  3. Stacks A stack is a Last in First out (LIFO) data structure in which all insertions and deletions are restricted to one end called top. Figure 4-1

  4. Stack Basic Operations Overflow? Figure 4-2

  5. Stack Basic Operations Underflow? Figure 4-3

  6. Stack Basic Operations Figure 4-4

  7. Figure 4-5

  8. Stack Data Structure Figure 4-6

  9. Stack Data Structure Figure 4-7

  10. Stack Operations Figure 4-8, Part I

  11. Stack Operations Figure 4-8, Part II

  12. Create Stack • algorithm createStack • Allocates memory for a stack head node from dynamic memory and returns its address to the caller. • Pre Nothing • Post Head node allocated or error returned • Return pointer to head node or null pointer if no memory • if (memory available) • allocate (stackPtr) • stackPtrcount = 0 • stackPtrtop = null • else • stackPtr = null • return stackPtr end createStack

  13. Push Stack Figure 4-9

  14. Push Stack • algorithm pushStack(val stack <head pointer>, • val data <data type>) • Insert (push) one item into the stack. • Pre stack is a pointer to the stack head structure. • data contains data to be pushed into stack. • Post data have been pushed in stack. • if (stack full) • success = false • else • allocate (newPtr) • newPtrdata= data • newPtrnext= stacktop • stacktop=newPtr • stackcount = stackcount+1 • Success=true • return success end pushStack

  15. Pop Stack Figure 4-10

  16. Pop Stack • algorithm popStack(val stack <head pointer>, • val dataOut <data type>) • Pops the item on the top of the stack and returns it to the user. • Pre stack is a pointer to the stack head structure. • dataOut is a reference variable to receive the data. • Post data have been returned to the calling algorithm. • if (stack empty) • success = false • else • dltPtr= stacktop • dataOut = stacktopdata • stacktop = stacktopnext • stackcount = stackcount – 1 • recycle(dltPtr) • success=true • return success end popStack

  17. Destroy Stack • algorithm destroyStack(val stack <head pointer>) • This algorithm releases all nodes back to the dynamic memory. • Pre stack is a pointer to the stack head structure. • Post stack empty and all nodes recycled • if (stack not empty) • loop • temp = stacktop • stacktop = stacktoplink • recycled(temp) • recycled (stack) • return null pointer end destroyStack

  18. Stack Applications • We can be classified stack applications into four categories: • Reversing data. • Parsing data. • Postponing data usage. • Backtracking steps.

  19. Stack ApplicationsReversing Data 1 2 3 4 5 5 4 3 2 1 Push 5 Pop 4 3 2 1

  20. Stack ApplicationsParsing Data Breaks the data into independent pieces for further processing. source program machine language Compiler Figure 4-11

  21. Stack ApplicationsParsing Data 3.Pop ) 2.Push ( 1.Push ( 3.Pop ) 2.Pop ) 1.Push (

  22. Stack ApplicationsPosponement • Arithmetic expression can be represent three different formats: • Prefix + a b • Infix a + b • Postfix a b + Arithmetic precedence; multiply and divide before add and subtract!

  23. Stack ApplicationsPosponement • A + B * C • Place the paranthesis • ( A + ( B * C)) • Replace it in postfix format • (A(BC*)+) • Remove all paranthesis • ABC*+ Implementation by computer is too hard!

  24. Stack ApplicationsPosponement * and / operators have higher priority than the operator at the top of the stack. Figure 4-12, Part I

  25. Stack ApplicationsPosponement Figure 4-12, Part II

  26. Stack ApplicationsEvaluation of Postfix Expression Figure 4-13

  27. Excercise • Change the following infix expression to postfix expression using the algoritmic method (a stack). a+b*c-d Solution: abc*+d-

  28. Infix String : a+b*c-d • The first character scanned is 'a'. 'a' is added to the Postfix string. The next character scanned is '+'. It being an operator, it is pushed to the stack. • Next character scanned is 'b' which will be placed in the Postfix string. Next character is '*' which is an operator. Now, the top element of the stack is '+' which has lower precedence than '*', so '*' will be pushed to the stack. • The next character is 'c' which is placed in the Postfix string. Next character scanned is '-'. The topmost character in the stack is '*' which has a higher precedence than '-'. Thus '*' will be popped out from the stack and added to the Postfix string. Even now the stack is not empty. Now the topmost element of the stack is '+' which has equal priority to '-'. So pop the '+' from the stack and add it to the Postfix string. The '-' will be pushed to the stack. • Next character is 'd' which is added to Postfix string. Now all characters have been scanned so we must pop the remaining elements from the stack and add it to the Postfix string. At this stage we have only a '-' in the stack. It is popped out and added to the Postfix string. So, after all characters are scanned, this is how the stack and Postfix string will be : • End result : • * Infix String : a+b*c-d • * Postfix String : abc*+d-

  29. Stack ApplicationsBackTracking Figure 4-14

  30. Figure 4-15

  31. Stack ApplicationsBackTracking Figure 4-16

  32. Figure 4-17

  33. Array Implementation of Stacks stack stackAry <pointer to array of dataType> count <integer> stackMax <integer> top <index> end stack Figure 4-20

  34. Hw-5 • Write a program that accepts parentheses and brackets characters, one per line on standard input. Use a stack to determine whether pairs of characters are matching or not. Therefore complete the body of below function. • bool balanced(const char p[ ], size_t n) • // Precondition: p[0]...p[n-1] contains n characters, each of which • // is '(', ')', '{' or '}'. • // Postcondition: The function returns true if the characters form a • // sequence of correctly balanced parentheses with each '(' matching • // a ')' and each '{' matching a '}'. Note that a sequence such as • // ( { ) } is NOT balanced because when we draw lines to match the • // parentheses to their partners, the lines cross each other. On the • // other hand, ( { } ) amd { ( ) } are both balanced. Load your HW-5 to FTP site until 13 Apr. 07 at 09:00 am.

  35. Exercises Projects – 23 page 211, 212 Figure 4-21

  36. Exercises Projects – 24 page 212 Figure 4-22

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