The Question is : Three stacks can be used to sort a list of numbers. Assuming stack in holds the input list of numbers, stack out is to hold the output list after sorting the numbers and temp is used during the sorting process. The sorting algorithm follows.
1 set up stack in and print it 2 while stack in is not empty repeat 2.1 max = in.pop 2.2 while there are still element in stack in repeat 2.2.1 value = in.pop 2.2.2 if value > max 2.2.2.1 temp.push(max) 2.2.2.2 max = value 2.2.3 else 2.2.3.1 temp.push(value) 2.3 in = temp 2.4 out.push(max) 2.5 temp.clear
this my coding
import java.util.*;
public class MainAssignment3 { public static void main(String[]args) { int Max =0; int Value =0;
LinkedList<Integer>Input=new LinkedList<Integer>(); LinkedList<Integer>Temp=new LinkedList<Integer>(); LinkedList<Integer>OutPut=new LinkedList<Integer>();
Input.addLast(90); Input.addLast(21); Input.addLast(33); Input.addLast(80); Input.addLast(67);
System.out.println("The Input Stack is : " + Input);
while(!Input.isEmpty()) { Max = Input.removeLast(); Value = Input.removeLast();
System.out.println("MAx: " +Max); System.out.println("Value: " +Value); if (Value > Max) { Temp.push(Max); Max=Value; } else { Temp.push(Value); } Input = Temp; OutPut.push(Max); Temp.clear();
} System.out.println("The Output Stack is: " + OutPut);
} }
the answer i get is
The Input Stack is : [90, 21, 33, 80, 67] MAx: 67 Value: 80
i want to get the sorted stack in Output stack
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