Stacks and Queues Pepper

Why History Simplicity Operating Systems – Function Call Stack

What is a Queue First In, First Out (FIFO) Standard queue methods – add() – remove() – isEmpty() – peek() – size() Interface in Java – Can create as a LinkedList

What is a Stack Last In, First Out (LIFO) Standard Stack Methods – push() – pop() – isEmpty() – peek() – size()

Creating and Adding Items Create both and add 5 items Stack st = new Stack ( ); Queue qu = new LinkedList (); for (int c = 0; c < 5; c++) { qu.add(Integer.toString(c)); st.push(Integer.toString(c)); }

Iterating through every item Code to just run through until empty: while (!st.isEmpty()){ System.out.println(st.pop()); } while (!qu.isEmpty()){ System.out.println(qu.remove()); } Empties the structure though

Queue: Iterate through all & Recreate Queue: Just control by initial size and keep adding back for (int c = 0; c < qu.size(); c++){ String value = qu.remove(); System.out.println(value); qu.add(value); } Removing values – Move the qu.size() call to before the loop because the size will change.

Stack: Iterate through all & Recreate Stack: Cannot just push back because it will become your next value. Need auxillary structure. Stack temp = new Stack (); while (!st.isEmpty()){ String value = st.pop(); System.out.println(value); temp.push(value); } while (!temp.isEmpty()){ st.push(temp.pop()); }

Read Stack – if you only had a Queue Read through the stack and place on queue – Queue holds reverse order Place queue on stack – still reverse Read through the stack and place on queue – Queue now holds original order Place queue on stack – original order

Compare two stacks using one temp Compare every item of Two stack contents Leave stacks the same when done When one has more items, it cannot be the same When you find a false result, cannot just return false – have to put the items back What order to put items back? How do you get a true result?

Comparing stacks public static boolean isSameStack(Stack st1, Stack st2){ Stack temp = new Stack (); boolean res = true; if (st1.size() != st2.size()) { return false; } else { while (!st1.isEmpty()){ String value1 = st1.pop(); String value2 = st2.pop(); temp.push(value1); temp.push(value2); if (!value1.equals(value2)){ // break and set value to false; res = false; break; } } // now put all the values back: while (!temp.isEmpty()){ st2.push(temp.pop()); st1.push(temp.pop()); } return res; // it will be true if nothing ever set it to false; } }

Splitting Tokens To split an expression like (4 + (5 – 3)) Why not Scanner? Making your own splitter – Holds Queue of characters Next entry waiting Special characters we are searching for – Methods peek hasNext next

Loading the Character Queue public StringSplitter(String str){ ch = new LinkedList (); for (int c = 0; c < str.length(); c++){ ch.add(str.charAt(c)); } findNextToken(); }

How to find the next token Skip whitespace If the queue is empty, set the token to null Pull out one character – If it is a special character, you are done – If it is a number, keep going until you get whitespace or a special character

Finding the Token public String findNextToken(){ this.token = ""; // skip any whitespace while (!ch.isEmpty() && Character.isWhitespace(ch.peek())){ ch.remove(); // just throw away the whitespace } // now we are past the whitespace. Continue until end or special char //if the next one is empty there is no token if (ch.isEmpty()) { token = null; } else // something in token {

Finding the token part 2 token = "" + ch.remove(); //was it a special character or number if (!SPECIAL_CHAR.contains(token)){ // keep going until end of number skipping white space, // ending at number's end boolean done = false; while (!ch.isEmpty() && !done){ char nextone = ch.peek(); if (Character.isWhitespace(nextone) || SPECIAL_CHAR.contains("" + nextone)) { done = true;} else { token = token + ch.remove(); } } } } return token; }

Next The token is already waiting for you, so give it and determine the next token. public String next(){ if (token == null) { throw new NoSuchElementException(); } String nextToken = token; findNextToken(); // reset the token return nextToken; }

Simple test public class StringSplitterTester { public static void main (String[] args) { StringSplitter ss = new StringSplitter( "18.4-((2.3*8.5 ) / ( (2.7 ^ 4.9 ))))"); while(ss.hasNext()){ System.out.println(ss.next()); }

Evaluating the Expression Dijkstra's method Push all symbols on one stack All numbers on the other stack When you reach a ), evaluate 2 numbers and their one operation Push that result back onto the stack

Expression – Create the ADTs Stack symbols = new Stack (); Stack numbers = new Stack (); StringSplitter ss = new StringSplitter(expression); boolean bad = false;

Evaluate Expression - body while(ss.hasNext()){ String token = ss.next(); if (token.equals(")")) { // process the expression } else if (StringSplitter.SPECIAL_CHAR.contains(token)) { symbols.push(token); } else { numbers.push(Integer.parseInt(token)); } } if (bad) { System.out.println("The answer is - no answer"); } else { System.out.println("The answer is " + numbers.pop()); } } }

Calculating 2 operands String operation = symbols.pop(); Integer firstNum = numbers.pop(); Integer nextNum = numbers.pop(); String openparen = symbols.pop(); if (openparen.equals("(")) { if (operation.equals("+")){ numbers.push(firstNum + nextNum); } else if (operation.equals("-")) { numbers.push( nextNum - firstNum); } else { bad = true; } } else { bad = true; }

HTML Splitter Split by instead of (*,etc Toss anything not inside Need to identify the starter and ender characters: – < – >

HTML Processor Read through the queue pushing onto the stack whenever a token must be matched. – When you encounter a / at the end, it means it is a self closing tag, so don't push it onto the stack When you encounter a / at the beginning, it is supposed to close something that is right before it. – Pop the last item, it should have matching first words of the tag.

Stacks and Queues What Why Patterns: – Read and restore Queues just keep moving onto itself Stacks require an auxillary storage – Compare