交通大學資訊工程學系 Programming in Java More examples, … 蔡文能交通大學資訊工程學系 [email protected] tsaiwn/java//java

交通大學資訊工程學系

Programming in Java

More examples, …

蔡文能交通大學資訊工程學系[email protected]

http://www.csie.nctu.edu.tw/~tsaiwn/java/

交通大學資訊工程學系蔡文能 7-第 2頁

Java More Java Examples

Agenda

Computing factorial Loop method Recursive method

Computing primes Sieve method

Sorting an array Selection Sort Insertion Sort Bubble Sort Quick Sort Merge Sort



Computing FactorialComputing Factorial• The factorial of an integer is the product of that number and all

of the positive integers smaller than it.- 0! = 1- 1! = 1

. . .- 5! = 5*4*3*2*1 = 120

- 6! = 6*5*4*3*2*1 = 720 . . .

- 50! =30414093201713378043612608166064768844377641568960512000000000000



Factorial -- IterationFactorial -- Iterationpublic class Factorial {

public static long factorial(int n) {long fact = 1;for (int i =2; i <= n; i ++) //for Loop

fact *= i; //shorthand for fact=fact*i;

return fact;}

}

public class ComputingFactorial {public static void main(String arg[ ]) {

long a = Factorial.factorial(Integer.parseInt(arg[0]));System.out.println("Factorial of " + arg[0] + "=" + a);

} }

In separate files



Recursion 遞迴概念很久很久以前 , 這裡有一座山 , 山裡有一座廟 , 廟裡有個老和尚和小和尚 , 小和尚要老和尚說故事給他聽。老和尚就說 :『很久很久以前 , 這裡有一座山 , 山裡友一座廟 , 廟裡有個老和尚和小和尚 , 小和尚要老和尚說故事給他聽。 .....莊子與惠子游於濠梁之上。莊子曰：「鯈於出游從容，是魚之樂也。」

惠子曰 :『子非魚焉知魚之樂 ?』莊子曰 :『子非我焉知我不知魚之樂 ?』惠子曰 :『子非我焉知我不知你不知魚之樂 ?』莊子曰 :『子非我焉知我不知你不知我知魚之樂 ?』惠子曰 :『子非我焉知我不知你不知我不知你不知魚之樂 ?』

莊子曰 : 『子非我焉知 ... !@#$%^&*? ...



Factorial – Recursion, in JavaFactorial – Recursion, in Java

/*** This class shows a recursive method to compute factorials. This method* calls itself repeatedly based on the formula: n! = n* (n-1)!**/public class Factorial2 {

public static long factorial(int n) {if (n == 1) return 1;else return n*factorial(n - 1);

}}



N Factorial ( Recursive 版 )

long factorial(int n) { if( n < 0) return -factorial(-n); if( n ==1 || n==0) return 1; return n * factorial(n-1);} /* 告訴我 n-1 階乘 , 我就給你 n 階乘 */#include <stdio.h>int main( ){ printf("5! = %ld\n", factorial(5));}

N 階乘就是 N 乘以 N-1 階乘

C語言版本



GCD -- 歐幾里得的最大公約數

輾轉相除法 (recursive概念 !)

GCD(m, n) = GCD(n, m%n)

但如果 n 是 0 則答案為 m

long gcd(long m, long n) {

if(n==0) return m;

return gcd(n, m%n);

}

如何寫成 non-recursive version ?



GCD non-recursive version

long gcd(long m, long n) { int r; /* remainder */ while(n != 0) { r = m%n; m = n; n = r; } return m;}



Recursive 也可以很有趣 (1/3)

Fibonacci Series: (費氏數列 )一開始有一對兔子小兔子隔一個月可以長大為成兔每對成兔每隔一個月可生出一對兔子假設兔子永遠不死 , 問第 n 個月時有幾對兔子？

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, . fib(n)0 1 2 3 4 5 6 7 8 9 10 11 = n




long fib(int n) { if( n < 0) return 0; if( n ==1 || n==0) return 1; return fib(n-1)+fib(n-2);} /* 給我 n 我就告訴你第 n 個月時有幾對兔子 */ /*第 n個月兔子數 = 第 n-1 個月兔子 + 第 n-2 個月兔子 */

/*但是 , 最開始兩個月例外 ! */




Fibonacci Series(費氏數列 ):1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, …

34/55 = 55/89 = 89/144 = 0.618 黃金分割比 :)

神奇的費氏數列 : 任何事物接近這些數字會有變化

請看 . .可怕的巧合 : !? 三重魔力 . . .

民國 34 年台灣光復 , 民國 89年變天國民黨從日本手上搶回台灣執政剛好 55 年 !



問題與思考 (Recursion)

寫出 non-recursive 版的 Fibonacci function ?

考慮小兔子隔兩個月可以長大為成兔 ?

考慮成兔的懷孕期是兩個月 ?

其他 Recursive 問題 Recursive 版的 binary search Quick sort Hanoi tower problem …



Factorial – cache ans in a TableFactorial – cache ans in a Table public class Factorial3 {

//create an array to cache values 0! Through 20!static long[ ] table = new long[21];static {ans[0] = 1;} //factorial of 0 is 1

static int last = 0; public static long factorial(int n) { if(n <= last) return ans[n];

long tmp = ans[last]; int k = last; while (k < n) { tmp = (k+1) * tmp; /* (k+1) ! */ if(last < 20) {++last; ans [last ] = tmp;} ++k;

} /* while */ return tmp;

} } Cache 唸作 cash



Computing Primes (1/3)Computing Primes (1/3)

• Finding the largest prime number smaller than a specified integer:Input integer m, find p m such that p is a prime and if there is prime p’ > p then p’ must be larger than m.

• Finding all prime numbers that smaller than a specified integer?

11 22 4433 55 7766 88 99 1212 1414 1515 1616 1818 20201010 1111 1313 1717 1919




• Algorithmmain idea: find primes by eliminating multiples of the form k j, where j is a prime smaller than square-root(m) and k is an integer such that k j m.

...... ......primeprime22 ii jj square-root(m)

223344

......

...... ......

22

33

44

22

33

44

22

22

22

ii

ii

ii

jj

jj

jj




• Finding all prime numbers that smaller than a specified integer?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20



Example: Find Primes (1/2)Example: Find Primes (1/2)

Import java.lang.*;

public class Sieve {public static void main(String[] args) {

int max = 100; //Assign a default valuetry {max = Integer.parseInt(arg[0]);}catch (Exception e) {} //Silently ignore exceptions.

//Create an array that specifies whether each number is prime or not.Boolean[] isprime = new boolean[max+1];

//Assume that all numbers are primes, until proven otherwise.for (int i = 0; <= max; i++) isprime[i] = true;

//We know that that 0 and 1 are not prime. Make a note of it.isprime[0] = isprime[1] = false;



Example: Find Primes (2/2)Example: Find Primes (2/2) //To compute all primes less than max, we need to rule out multiples of all //integers less than the square root of max.

int n = (int) Math.ceil(Math.sqrt(max));

for (int i = 0; i <= n; i++) {if (isprime[i]) { int k = 2;

for (int j = k*i; j <= max; j = (k ++)*i) isprime[j] = false; }

}

int largest;for (largest = max; !sprime[largest]; largest--); //empty loop bodySystem.out.println("The largest prime less than or equal to " + max + " is "

+ largest); }}



Sorting NumbersSorting Numbers

• Sorting:Input n numbers, sort them such that the numbers are ordered increasingly.

3 9 1 6 5 4 8 2 10 7

1 2 3 4 5 6 7 8 9 10



Selection Sort (1/4)Selection Sort (1/4)

• A simple sorting algorithmPurpose:

Input: an array A containing n integers. Output: sorted array.

1. i := 2;2. Find the least element a from A(i) to A(n);3. If a is less than A(i - 1), exchange A(i - 1) and a;4. i := i + 1; if (i <= n) goto step (2).




•

1st step: 3 9 1 6 5 4 8 2 10 7

2nd step: 1 9 3 6 5 4 8 2 10 7

1 2 3 6 5 4 8 9 10 7… ...

swap

swap



Selection Sort (3/4)Selection Sort (3/4)• Sorting program:

public class SelSort {public static void sort(double[ ] nums) {

for(int i = 0; i < nums.length; i++) {int min = i +1;for (int j = i+2; j < nums.length; j++) {

if (nums[j] < nums[min]) min = j;}if (nums[i] > nums[min]) {

double tmp;tmp = nums[i]; nums[i] = nums[min]; nums[min] =

tmp; }

} } }




• In separate file:

public class SortNumber {public static void main (String[ ] args) {

double[ ] nums = new double[10]; //Create an array to hold numbers

for(int i = 0; i < nums.length; i++) //Generate random numbers nums[i] = Math.random( )*100;

SelSort.sort(nums); //Sort themfor (int j = 0; j < nums.length; j++) //Print them out

System.out.println(nums [j] );}

} Math.random( ) : see next slide



Insertion SortInsertion Sortpublic class InsSort {

public static void sort(double[ ] nums) {for(int i = 1; i < nums.length; i++) {

double tmp = nums[i]; /* we want to put this in proper position */ k = i – 1;

while ((k>=0) && nums[k] > tmp ) { /* smaller than me */ nums[k+1] = nums[k]); /* move it down */

--k;}

nums[k+1] = tmp; /* copy it into correct position */ }

} }



Test the Insertion SortTest the Insertion Sort

• In separate file as previous example (Selection sort):

public class SortNumber {public static void main (String[ ] args) {

double[ ] nums = new double[10]; //Create an array to hold numbers

for(int i = 0; i < nums.length; i++) //Generate random numbers nums[i] = Math.random( )*100;

InsSort.sort(nums); //Sort them using insertion sort

for (int j = 0; j < nums.length; j++) //Print them outSystem.out.println(nums [j] );

}}



Bubble SortBubble Sortpublic class BubSort {

public static void sort(double[ ] nums) { for(int i = nums.length-2; i >=0; --i) {

int flag = 0; /* Assume no exchange */ for(k=0; k <= i; ++k) { /* walk through all every pass */

if ( nums[k] > nums[k+1] ) { /* incorrect order */ double tmp=nums[k]; nums[k]=nums[k+1]; nums[k+1]=tmp; ++flag; }

} if(flag == 0) break; /* no need to do next pass */

} } }



Algorithm quick_sort(array A, from, to)Input: from - pointer to the starting position of array A

to - pointer to the end position of array AOutput: sorted array: A’

1. Choose any one element as the pivot; 2. Find the first element a = A[i] larger than or equal to pivot from

A[from] to A[to];3. Find the first element b = A[j] smaller than or equal to pivot from

A[to] to A[from]; 4. If i < j then exchange a and b;5. Repeat step from 2 to 4 until j <= i;6. If from < j then recursive call quick_sort(A, from, j);7. If i < to then recursive call quick_sort(A, i, to);

Quick Sort (1/6)



• Quick sortmain idea:

1st step: 3 1 6 5 4 8 10 7

2nd step: 3 2 1 5 8 9 10 7

3rd step: 3 2 1 4 5 6 8 9 10 7

Choose 5 as pivotfrom to

2299

66 44

Smaller than any integerSmaller than any integerright to 5right to 5

greater than any integergreater than any integerleft to 5left to 5

ii jj

Quick Sort (2/6)



• Quick sort

4th step: 2 4 5 6 10 9

5th step: 1 2 3 4 5

pivotfrom to

33 11 88 77

5 6 7 8 10 95 6 7 8 10 96th step:6th step:

pivotfrom to

7 8 10 97 8 10 97th step:7th step:

9 109 108th step:8th step:

Quick Sort (3/6)



public class QuickSorter {public static void sort (int[ ] a, int from, int to) {

if ((a == null) || (a.length < 2)) return;int i = from, j = to;int pivot = a[(from + to)/2];do {

while ((i < to) && (a[i] < pivot)) i++;while ((j > from) && (a[j] >= pivot)) j--;if (i < j) { int tmp =a[i]; a [i] = a[j]; a[j] = tmp;}i++; j--;

}while (i <= j);if (from < j) sort(a, from, j);if (i < to) sort(a, i, to);

}}

Quick Sort (4/6)



14

3, 4, 6, 1, 10, 9, 5, 20, 19, 141, 12, 2, 15, 21, 13, 18, 17, 8, 16,

3, 4, 6, 1, 10, 9, 5, 8, 19, 1, 12, 2, 15, 21, 13, 18, 17, ,16,

3, 4, 6, 1, 10, 9, 5, 8, 13 ,1, 12, 2, 15, 21, 19, 18, 17, 20, 16,

i

j3, 4, 6, 1, 10, 9, 5, 8, 13 , 1, 12, 2

ii jj

Quick Sort (5/6)3, 4, 6, 1, 10, 9, 5, 20, 19, 14, 12, 2, 15, 21, 13, 18, 17, 8, 16, 1

jjii20,

14

ii jj3, 4, 6, 1, 10, 9, 5, 8, 13 , 1, 12, 2, 14, 21, 19, 18, 17, 20, 16, 1514



public class BQSorter {public static void sort (int[ ] a, int from, int to) {

if ((a == null) || (a.length < 2) || from >= to) return;int k = (from + to)/2; int tmp =a[to]; a [to] = a[k]; a[k] =

tmp;int pivot = a[to]; int i = from, j = to-1; while(i < j ) {

while ((i < j) && (a[i] < pivot)) i++;while ((i < j) && (a[j] >= pivot)) j--;if (i < j) { tmp =a[i]; a [i] = a[j]; a[j] = tmp;}

}; tmp =a[i]; a [i] = a[to]; a[to] = tmp;

if (from < i-1) sort(a, from, i-1);if (i < to) sort(a, i+1, to);

}}

Quick Sort (6/6)



Merging means the combination of two or more ordered sequence into

a single sequence. For example, can merge two sequences: 503, 703, 765

and 087, 512, 677 to obtain a sequence: 087, 503, 512, 677, 703, 765.

A simple way to accomplish this is to compare the two smallest items,

output the smallest, and then repeat the same process.

503 703 765087 512 677 087

503 703 765512 677

087 503703 765512 677

Merge Sort (1/3)



Algorithm Merge(s1, s2)Input: two sequences: s1 - x1 x2 ... xm and s2 - y1 y2 ... yn

Output: a sorted sequence: z1 z2 ... zm+n.1.[initialize] i := 1, j := 1, k := 1;2.[find smaller] if xi yj goto step 3, otherwise goto step 5;3.[output xi] zk.:= xi, k := k+1, i := i+1. If i m, goto step 2;4.[transmit yj ... yn] zk, ..., zm+n := yj, ..., yn. Terminate the

algorithm;5.[output yj] zk.:= yj, k := k+1, j := j+1. If j n, goto step 2;6.[transmit xi ... xm] zk, ..., zm+n := xi, ..., xm. Terminate the

algorithm;

Merge Sort (2/3)



Algorithm Merge-sorting(s)

Input: a sequences s = < x1, ..., xm>Output: a sorted sequence.1. If |s| = 1, then return s;2. k := m/2;3. s1 := Merge-sorting(x1, ..., xk);4. s2 := Merge-sorting(xk+1, ..., xm);5. return(Merge(s1, s2));

Merge Sort (3/3)



Time complexity of Mergesort

Takes roughly n·log2 n comparisons.

Without the shortcut, there is no best or worst case.

With the optional shortcut, the best case is when the array is already sorted: takes only (n-1) comparisons.



Binary Search (1/5)

Works for an array of numbers or objects that can be compared and are arranged in ascending (or descending) order.

A very fast method: only 20 comparisons are needed for an array of 1,000,000 elements; (30 comparisons can handle 1,000,000,000 elements; etc.)



Binary Search (2/5)Main idea: “divide and conquer”: compare the target value to the middle element; if equal, all set if smaller, apply binary search to the left half; if larger, apply binary search to the right half.

... MI MN MO MS MT NC ND ...

... MI MN MO MS MT NC ND ...

NJv

NJ



Binary Search (3/5)Recursive implementation:

public int binarySearch (int arr [ ], int value, int left, int right){ int middle = (left + right) / 2;

if (value == arr [middle] ) return middle;

else if ( value < arr[middle] && left < middle ) return binarySearch (arr, value, left, middle - 1);

else if ( value > arr[middle] && middle < right ) return binarySearch (arr, value, middle + 1, right );

else return -1; // Not found}



Binary Search (4/5)Iterative implementation:

public int binarySearch (int arr [ ], int value, int left, int right){ while (left <= right) { int middle = (left + right) / 2;

if ( value == arr [middle] ) return middle; else if ( value < arr[middle] ) right = middle - 1; else /* if ( value > arr[middle] ) */ left = middle + 1; } return -1; // Not found}



Binary Search (5/5)

The average number of comparisons is roughly

log2 n

The worst-case is

log2 (n + 1)

rounded up to an integer (e.g. 2 for n = 3, 3 for n = 7, 4 for n = 15, etc.)



PRNGPseudo Random Number Generator

java.lang.Math.random( ) A real (double) number: Uniform random number [0, 1)

java.util.Random A utility class in java.util package Has many miscellaneous random # generating functions java.lang.Math.random( ) actually uses nextDouble( ) in this class nextInt( ) returns a number like int rand( ) in C language nextGaussian( ) returns a # in standard Normal distribution



java.util.Random (1/2)

Benchmarks uses the java.util.Random class — a more controlled way to generate random numbers.

Constructors:

If we set the same “seed,” we get the same “random” sequence.

Random generator = new Random();

Random generator2 = new Random(seed);

Default “seed”: System.currentTimeMillis()

long seed;



java.util.Random (2/2)Methods:

int k = generator.nextInt ();

int k = generator.nextInt (n);

double x = generator.nextDouble ();

All 232 possible int values are produced with (approximately) equal probability

0 k < n

0 x < 1



Normal Distribution Random number



java.util.Arrays

Provides static methods for dealing with arrays.

Works for arrays of numbers, Strings, (and “comparable” Objects).

Methods:

int pos = Arrays.binarySearch (arr, target);

Arrays.sort (arr);Arrays.sort (arr, from, to);

Arrays.fill (arr, value); // fills arr with a given valueArrays.fill (arr, from, to, value);



java.util.List Interface

The List library interface describes a list of objects in abstract terms

In a list, objects are arranged in sequence

obj0, obj1, ..., objn-1

In Java, a list holds references to objects

A list can contain duplicate objects

(both obj1.equals(obj2) and obj1 == obj2 )



List Methods (a Subset)

int size();

boolean isEmpty ();

boolean add (Object obj);

void add (int i, Object obj);

Object set(int i, Object obj);

Object get(int i);

Object remove(int i);

boolean contains(Object obj);

int indexOf(Object obj);

returns true

inserts obj as the i-th value; i must be from 0 to size()

i must be from 0 to size() -1

use equals to compare objects



java.util.ArrayList (1/6)

Implements List using an array

Keeps track of the list capacity (the length of the allocated array) and list size (the number of elements currently in the list)

"Cat" "Hat" "Bat"

capacitysize



ArrayList (2/6)

Automatically increases (doubles) the capacity when the list runs out of space; allocates a bigger array and copies all the values into it

get(i) and set(i, obj) are efficient because an array provides random access to its elements

Throws IndexOutOfBoundsException wheni < 0 or i size()



ArrayList (3/6)

ArrayList holds objects (of any type)

If you need to put ints or doubles into a list, use a standard Java array or convert them into Integer or Double objects

You have to remember what types of objects your put into the list and may need to cast a retrieved object back into its type



ArrayList (4/6)From Java API Docs:

JDK1.5: ArrayList(Collection<? extends E> c) Constructs a list containing the elements of the specified collection, in the order they are returned by the collection's iterator.



ArrayList list = new ArrayList (); list.add (new Integer(1)); list.add (new Integer(2)); list.add (new Integer(3));

int sum = 0; for (int i = 0; i < list.size(); i++) sum += ((Integer) list.get(i)) . intValue();

ArrayList (5/6)Example 1

Need a cast to Integer in order to call intValue



ArrayList words = new ArrayList (4); words.add ("One"); words.add ("Fish"); words.add ("Two"); words.add ("Fish"); int i = 0;

while (i < words.size() ) { if ( ”Fish".equals (words.get(i)) ) words.remove(i); else i++; }

ArrayList (6/6)Example 2

Shifts all the values after the i-th to the left and decrements the size



Generic type



Generic ProgrammingDefine software components with type parameters A sorting algorithm has the same structure, regardless of the types

being sorted Stack primitives have the same semantics, regardless of the objects stored on the stack.

Compile time parameters

Implementation Same code used for different parameter values Different code generated for different values

C++ model: template class

Java JDK1.5: type inference



Exception, More examples, …

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交通大學資訊工程學系 Programming in Java More examples, … 蔡文能 交通大學資訊工程學系 [email protected] tsaiwn/java//java

交通大學資訊工程學系 Programming in Java More examples, … 蔡文能交通大學資訊工程學系 [email protected] tsaiwn/java//java