CS1101 Group1
description
Transcript of CS1101 Group1
CS1101 Group1
Discussion 5
Lek Hsiang Hui
lekhsian @ comp.nus.edu.sg
http://www.comp.nus.edu.sg/~lekhsian/cs1101
Debugging in Dr Java
Checklist• Setting a breakpoint
– Before running program– During the debugging process
• Step into a method• Watches
Debugging in Dr Javapublic class Dummy{ public static void main(String[] args){ int tmp = 0; for (int i = 0 ; i < 10 ; i++){ tmp++; } dummyMethod(); } public static void dummyMethod(){ System.out.println("do nothing1"); System.out.println("do nothing2"); System.out.println("do nothing3"); }}
• Ctrl+B (toggle breakpoints at current line)• F7 (Resume/Continue Run)• F11 (Step Over)• F12 (Step Into)• Shift + F12 (Step Out)
Search And Sorting
Searching
• Linear Search– Go through all the items in an array
• Binary Search– Search by “cutting” the possible solutions by
½ each time
Sorting
• Bubble Sort
• Insertion Sort
• Selection Sort
Stable Sorts
• A stable sort is one where the relative ordering of elements with the same value is preserved after sorting.– Two elements having the same key appear in the same order in
the sorted sequence as they did in the original sequence.
• Example:– Stable sort:
Before sorting: 6 2 3a 8 3b 5 9
After sorting: 2 3a 3b 5 6 8 9
– Unstable sort:
Before sorting: 6 2 3a 8 3b 5 9
After sorting: 2 3b 3a 5 6 8 9
• Which of the three basic sorts – bubble sort, selection sort, insertion sort – is/are stable?
Slide taken from cs1101x lecture notes
Sorting Algorithms Comparison
Name Best Time Worst Time Stable(in general)
Bubble sort O(n) O(n2) Yes
Insertion sort O(n) O(n2) Yes
Selection sort
O(n2) O(n2) No
Sorting Algorithms Comparison
Name Best Time Worst Time Stable(in general)
Bubble sort about
n comparisons
about
n2 comparisons
Yes
Insertion sort about
n comparisons
about
n2 comparisons
Yes
Selection sort about
n2 comparisons
about
n2 comparisons
No
Sorting Question
Which of the 3 algorithms is best for data
that is almost sorted?
Recursion
Recursion
• Base case
• Recursive case
n! = n * (n-1) * (n-2) * … * 2 * 1 for n > 0 (recursive case)
0! = 1 (base case)
int fac(int n){
if (n == 0) return 1; else{ return n * fac(n-1); }
}
Recursion
• General approaches
• Think recursive– Split the problem into simple cases– Work in the opposite way
• Work from a base case• See what you need to do for the 2nd last case• etc
Recursion Practice
• Can you write a “while” method that does what the while loop is doing
Recursion Practice
• Can you write a “while” method that does what the while loop is doing
public static void While (int i){
if (cond){
//code of while body
While(i);
}
}
Recursion Practice
while(i > 0){ System.out.println(i); i--;}
public static void While (int i){ if (i > 0 ){ System.out.println(i); i--; While(i); } }
References
• Pass by Value– Java is always pass by value!
• Pass by Reference
Pass by Valuepublic static void main (String args[]){
//a is an Integer Object
//you can just think of it as a
//integer value
Integer a = new Integer(13);
addOne(a);
System.out.println(a);
}
public static void addOne(Integer number){
number = new Integer(number.intValue() + 1); System.out.println("number inside the method : " + number);
}
What do you think is the output?
Pass by Valuepublic static void main (String args[]){
//a is an Integer Object
//you can just think of it as a
//integer value
Integer a = new Integer(13);
addOne(a);
System.out.println(a);
}
public static void addOne(Integer number){
number = new Integer(number.intValue() + 1); System.out.println("number inside the method : " + number);
}
What do you think is the output?
number inside the method : 14
13
Pass by Valuepublic static void main (String args[]){
Integer a = new Integer(13);
addOne(a);
System.out.println(a);
}
public static void addOne(Integer number){…}
What is really happening on the JVM?
13
0x0010
The value of a is at memory location 0x0010
Note that this is really a simplified view of what’s happening
Pass by Valuepublic static void main (String args[]){
Integer a = new Integer(13);
addOne(a);
System.out.println(a);
}
public static void addOne(Integer number){…}
What is really happening on the JVM?
13
0x0010
0x0010
a
0x0020
To find the value of a, just need to go to
memory location 0x0010This information is stored at
memory location 0x0020
Pass by Valuepublic static void main (String args[]){
Integer a = new Integer(13);
addOne(a);
System.out.println(a);
}
public static void addOne(Integer number){…}
What is really happening on the JVM?
13
0x0010
0x0010 addOne(a)
What happens when you call a method is that another box with the value 0x0010 is passed in
(both are pointing to 0x0010)
What this means is that you can, only change the contents
at 0x0010, but you cannot define anew totally new object for a
(You are allowed to modify 0x0010)a
0x0020 0x0010
Copy of a
0x0030
Pass by Referencepublic static void main (String args[]){
Integer a = new Integer(13);
addOne(a);
System.out.println(a);
}
public static void addOne(Integer number){…}
What is really happening on the JVM?
13
0x0010
0x0010 addOne(a)
In Pass by Reference,what happens call a method
is that the first box on the left is passed
to the method(0x0010)
What this means is that you can, change the contents
at 0x0010, and you can also give new pointer for Integer a
(You are allowed to modify both 0x0010, and 0x0020)
a
0x0020
Java (Pass by value)
• When you are passing primitive data types to methods (int, byte, double etc), you can think of it as you are giving the method the numerical value of the variable, not the variable itself
Java (Pass by value)
• When you are passing in references…– Objects, Arrays*– Can really think of Arrays are Objects
• You are given access to the content of that Object (but cannot modify the container)
Java (Pass by value)
int[] arr = new int[10];
method(arr);
public static void method(int[] a){
a = new int[20];
}
arrcontents
0x0010
0x0010
arr
0x0020 0x0010
Copy of arr
0x0030
Java (Pass by value)
int[] arr = new int[10];
method(arr);
public static void method(int[] a){
a = new int[20];
}
arrcontents
0x0010
0x0010
arr
0x0020 0x0050
Copy of arr
0x0030
new array
When you comes out of the
method, you don’t really have the
previous box and 0x0050
Java (Pass by value)
int[] arr = new int[10];
method(arr);
public static void method(int[] a){
a = new int[20];
}
arrcontents
0x0010
0x0010
arr
0x0020