![]() ![]() Iterating through Stack Items - Stack Iteratorīy definition of stack data structure you would use it as a container where items are added to and removed from one end. IsEmpty(): Returns true if stack is empty, false otherwise. Size(): Return the number of objects the stack contains right now. In addition to push() and pop() methods we can also define a few supporting but optional methods, such as, Pop(): Return the top object from the stack, and remove as well. The following methods we plan to implement as part of our stack implementation in Java using linked list. However, in linked implementation of stack we don't exactly require the top pointer because we will add an item at the beginning of the list and remove it from the beginning of the list.Ī stack by definition supports two methods, one is push for adding objects to the stack, and second, pop for removing the latest added object from the stack. ![]() To insert objects into and remove from stack a pointer usually called top is maintained that points to last inserted item. A stack is a container to which objects are added and removed by following last-in-first-out strategy. Linked list implementation of stack is efficient than array implementation because it does not reserve memory in advance. This article demonstrates a linked list implementation of generic stack. Iterating through Stack Items - Stack Iterator.These are the values that define a student. In your example, Students appear to have a name, an age, a course, a year, and a section. What could you ask it to do? You could ask it to tell you its name, to tell you whether it is enrolled in a given course or not, whether it shares a course with a given other student or not, etc etc. What properties might it have? For example, it may have a name, a gender, a major, a list of courses, etc etc. To start with OOP, you should think of the Student class as an Object, a tangible thing like a ball. We must catch a NumberFormatException in the event the user fails to enter an int, and in this case we'll just prompt them again. For our ints we'll use parseInt() to convert first. For our string values we can assign this directly. You saw it skip over name and go straight to age for a similar reason - the call to nextInt() captured your menu choice after the user hits enter but it did not capture the line return itself - the subsequent call to nextLine() did that (consuming both calls on the menuchoice and denying the user the chance to enter a name). Calling next() will return the next token, and supposing the user enters a series of \$n\$ words (a space delimited list of tokens) it will consume the next \$n\$ calls to next(). ![]() We want to call nextLine() to collect an entire line of input at a time. Student student = new Student(name, age, course, year, section) MenuChoice = Integer.parseInt(input.nextLine()) Īge = Integer.parseInt(input.nextLine()) ", course: " m_course ", year: " m_year Return "name: " m_name ", age: " m_age Public Student( String name, int age, String course, String year, String section ) This solution also uses an ArrayList to store the results, and a for-each loop to iterate through them. The interesting part is in main() and the use of Scanner. This was implemented independently from Msnik's answer (which is great by the way). It seems to skip and move on to ask the user for the age rather than starting on name. But in Java, it is different, as it won't let me input data on the string array name. Is there a better way to implement my code in Java? I used to code this in C and pretty much worked right off the bat. ![]()
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