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Stack Data Structure

Stack Data Structure. By : Imam M Shofi. What is stack?. A stack is a limited version of an array. New elements, or nodes as they are often called, can be added to a stack and removed from a stack only from one end.

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Stack Data Structure

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  1. Stack Data Structure By : Imam M Shofi

  2. What is stack? • A stack is a limited version of an array. • New elements, or nodes as they are often called, can be added to a stack and removed from a stack only from one end. • Access system a stack is referred to as a LIFO structure (Last-In First-Out) • Some illustrations: stack of CDs stack of satay

  3. Stacks operations • Push : adds a new node Push(X,S)  add the value X to the TOP of stack • Pop : removes a node Pop(S)  removes the TOP node and returns its value • IsEmpty : reports whether the stack is empty IsEmpty(S)  report whether the stack S is empty • IsFull : reports whether the stack is full IsFull(S)  report whether the stack S is full • Initialize : creates/initializes the stack Initialize(S)  create a new empty stack named S • Destroy : deletes the contents of the stack (may be implemented by re-initializing the stack) Destroy(S)  deletes the contents of the stack S

  4. Illustration/example Stack’s contents TOP value Operation 1. Initialiaze(S) <empty> 0 2. Push(‘a’,S) a 1 3. Push(‘b’,S) a b 2 4. Push(‘c’,S) a b c 3 5. Pop(S) a b 2 6. Push(‘d’,S) a b d 3 7. Push(‘e’,S) a b d e 4 8. Pop(S) a b d 3 9. Pop(S) a b 2 10. Pop(S) a 1

  5. Exercise • What would the state of the stack be after the following operations: create stack push A onto stack push F onto stack pop item from stack push B onto stack pop item from stack pop item from stack • Show the state of the stack and the value of each variable after execution of each of the following statements: A = 5 B = 3 C = 7 (a) create stack push A onto stack push C*C onto stack pop item from stack and store in B push B+A onto stack pop item from stack and store in A pop item from stack and store in B (b) create stack push B onto stack push C onto stack push A onto stack A=B*C push A+C onto stack pop item from stack and store in A pop item from stack and store in B pop item from stack and store in C

  6. Converting between notations • INFIX to PREFIX : (A+B) – (C*D) • Do the first brace: (A+B), the PREFIX is +AB • Do the second brace: (C*D), the PREFIX is *CD • The end is operator -: +AB - *CD, the PREFIX is -+AB*CD • INFIX to POSTFIX : (A+B) – (C*D) • Do the first brace: (A+B), the POSTFIX is AB+ • Do the second brace: (C*D), the POSTFIX is CD* • The end is operator -: AB+ - CD*, the PREFIX is AB+CD*- • PREFIX to INFIX : +/*A B C D • Find the first operator: *, take 2 operands before the operator (A and B), the INFIX is (A*B) • Find the second operator: /, take 2 operands before the operator (A*B and C), the INFIX is ((A*B)/C) • Find the third operator: +, take 2 operands before the operator (((A*B)/C) and D), the INFIX is ((A*B)/C)+D

  7. Converting between notations(2) • PREFIX to POSTFIX : +/*A B C D • Find the first operator: *, take 2 operands before the operator (A and B), the POSTFIX is AB* • Find the second operator: /, take 2 operands before the operator (AB* and C), the POSTFIX is AB*C/ • Find the third operator: +, take 2 operands before the operator (AB*C/ and D), the POSTFIX is AB*C/D+ • POSTFIX to INFIX : ABCD*/- • Find the first operator: *, take 2 operands before the operator (C and D), the INFIX is (C*D) • Find the second operator: /, take 2 operands before the operator ((C*D) and B), the INFIX is (B/(C*D) • Find the third operator: -, take 2 operands before the operator ((B/(C*D) and A), the INFIX is A – (B/(C*D) • POSTFIX to PREFIX : ABCD*/- • Find the first operator: *, take 2 operands before the operator (C and D), the PREFIX is *CD • Find the second operator: /, take 2 operands before the operator (*CD and B), the PREFIX is /B*CD • Find the third operator: -, take 2 operands before the operator (/B*CD and A), the PREFIX is -A /B*CD

  8. Exercise: Converting • Convert these INFIX to PREFIX and POSTFIX : • A / B – C / D • (A + B) ^ 3 – C * D • A ^ (B + C) • Convert these PREFIX to INFIX and POSTFIX : • + – / A B C ^ D E • – + D E / X Y • ^ + 2 3 – C D • Convert these POSTFIX to INFIX and PREFIX : • A B C + – • G H + I J / * • A B ^ C D + –

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