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They will scratch and mess up. They will scratch and mess up. They will scratch and mess up. They will scratch and mess up. They will scratch and mess up. They will scratch and mess up.

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Q: Which important property do stack algorithm have?

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Design an algorithm to show the different operations on a stack?

// stack to contain content Stack sourceStack = new Stack(); // ... fill sourceStack with content // stack to contain reversed content Stack targetStack = new Stack(); while (!sourceStack.empty()) { targetStack.push(sourceStack.pop()); } // targetStack contains the reversed content of sourceStack

Stock sorting algorithm is a algorithm which is used to sort any kind of stock i.e. any data type containing the primitive values like array ,link list ,stack etc.

Yes. It is possible to provide a solution to the diamond-square algorithm using Java and recursion.

One of the best property of an algorithm is that it is simple. It can not be too long and ambiguous, it has to be straightforward, with understandable variables as used also in a pseudocode.

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Design an algorithm to show the different operations on a stack?

Stack implementations allow us to easily implement backtracking algorithms.

A stack is a LIFO (last-in, first-out) data structure such that only the top-most element is accessible and all new elements are pushed onto the top (analogous to a stack of plates). Stacks are advantageous when implementing a back-tracking algorithm but are ultimately useless for anything else. However, this is not a disadvantage. If you're not implementing a back-tracking algorithm then the problem is not the stack itself it is the fact that you are using the wrong type of container for your algorithm.

// stack to contain content Stack sourceStack = new Stack(); // ... fill sourceStack with content // stack to contain reversed content Stack targetStack = new Stack(); while (!sourceStack.empty()) { targetStack.push(sourceStack.pop()); } // targetStack contains the reversed content of sourceStack

Stock sorting algorithm is a algorithm which is used to sort any kind of stock i.e. any data type containing the primitive values like array ,link list ,stack etc.

Stock sorting algorithm is a algorithm which is used to sort any kind of stock i.e. any data type containing the primitive values like array ,link list ,stack etc.

Yes. It is possible to provide a solution to the diamond-square algorithm using Java and recursion.

One of the best property of an algorithm is that it is simple. It can not be too long and ambiguous, it has to be straightforward, with understandable variables as used also in a pseudocode.

You overcome limitations of the stack in polygon filling, or in any other algorithm, far that matter, but using an iterative technique, rather than a recursive technique. Recursion is quite useful, and can simplify algorithm design. Polygon filling, however, is a class of algorithm can potentially have a very deep recursion depth. This causes stress on the stack, hence the need for iteration.

The belt-and-braces technique is easy enough: > > prefix_to_infix(stream, stack) > if stack is not empty > pop a node off the stack > if this node represents an operator > write an opening parenthesis to stream > prefix_to_infix(stream, stack) > write operator to stream > prefix_to_infix(stream, stack) > write a closing parenthesis to stream > else > write value to stream > endif > endif > endfunc

1. If TOP = MAXSTK THEN [Stack already filled?] Print "OVERFLOW" Go to step 4 Endif 2. TOP = TOP + 1 [Increase TOP by 1] 3. Set ST[STOP] = ITEM [Insert ITEM in new TOP position] 4. End

It is not possible to write a code to POP from the stack when there is no your stack implementation information.Because of that I am going to talk more about Stack in computer architecture and there will be additional link to specific examples(-e).In x86 architecture there is three registers (BP, SP and SS) which are connected with stack and only SP and SS is needed.SS - Stack Segment (base register);SP - Stack Pointer (offset);This is how the POP instruction works:# operand = [SS:SP] (top of the stack) # SP = SP + 2; (change SP to point to new top element)