'Algorithm&DataStructures'에 해당되는 글 38건

스택(Stack) - 간단자바예제(Simple in java)

1. 스택(Stack)

- Push : 탑에 객체가 쌓인다(add)

- Pop : 탑에 객체가 없어진다(remove)

- Peek : 탑의 값을 가져온다

2. 구현

1) 변수 초기화

private int stackSize; private int top; private int[] stackArray; public SimpleImplementation(int size){ this.stackSize = size; this.top = -1; this.stackArray= new int[stackSize]; }

2) push() and isStackFull()

public void push(int element) throws Exception{ if(this.isStackFull()){ throw new Exception("Stack is full"); } System.out.println("pushing:"+element); this.stackArray[++top] = element; } public boolean isStackFull(){ if(top == stackSize - 1){ return true; }else{ return false; } }  

3). pop() and isStackEmpty()

public int pop() throws Exception{ if(this.isStackEmpty()){ throw new Exception("Stack is empty"); } int element = stackArray[top--]; System.out.println("poping:"+element); return element; } public boolean isStackEmpty(){ if(top == -1){ return true; }else{ return false; } }  

4) peek()

public int peek(){ System.out.println("peek:"+stackArray[top]); return stackArray[top]; }  

5) main()

public static void main(String[]args){ SimpleImplementation stack = new SimpleImplementation(3); try{ stack.push(1); stack.push(2); stack.push(3); stack.pop(); stack.push(4); stack.peek(); stack.pop(); stack.pop(); stack.pop(); stack.pop(); }catch(Exception e){ System.out.println(e.getMessage()); } }  

OneNumber(한수)

1. Question    

- There is a integer X

- X has digit that is made of arithmetical sequence like 123, 468, 369

- You enter a number N( 1 <= N <=1000)

- To get a count of X within 1 <= N 

2. Solution

import java.util.Scanner; public class ArithmeticalSequence { void count(int a){ int count = 0; if(a <= 1000){ for(int i = 1; i<=a; i++){ int thousandsDigit = i/1000; int hundredDigit = i/100%10; int tenDigit = i/10%10; int unitDigit = i%10; int temp0 = thousandsDigit - hundredDigit; int temp = hundredDigit - tenDigit; int temp1 = tenDigit - unitDigit; if(thousandsDigit != 0){ if(temp0 == temp && temp==temp1){ count = count + 1; } }else{ if(hundredDigit != 0){ if(99<i){ if( temp == temp1){ count = count + 1; } } }else{ count = count + 1; } } } } System.out.println(count); } public static void main(String[]args){ ArithmeticalSequence cc = new ArithmeticalSequence(); Scanner aa = new Scanner(System.in); int b = aa.nextInt(); cc.count(b); aa.close(); } }

SelfNumber

1. Question

- If you start with 33, the next number is 33 + 3 +3 = 39, the next is 39 + 3 + 9 = 51 and so you generate the sequence

33, 39, 51, 57....

- d(n) n is called a generator of d(n)

- Some numbers have more than one generator

- A number with no generators is a self-number

- Write a program to output all positive self-numbers less than 10000 in increasing order, one per line

2. Solution

package stepByStep; public class SelfNum { int calc (int number) { int sum = 0; if (number < 10000){ int a = number / 1000; int b = number / 100 % 10; int c = number / 10 % 10; int d = number % 10; sum = a + b + c + d + number; } return sum; } public static void main (String[] args) { SelfNum aa = new SelfNum(); int arr[] = new int[10000]; int number = 1; for(int i = 0; i < 10000; i++, number++) { if (aa.calc(number) < 10000) { arr[aa.calc(number)] = aa.calc(number); } } for (int k = 1; k < arr.length; k++) { if(arr[k] == 0) { System.out.println(k); } } } }

SmallerThanX

1. Question

1) There is two numbers N and X in the first low

2) In the second low, You should enter A progression, Size of A progression will be N

3) To print out numbers that is smaller than X  

2. Solution

package stepByStep; import java.util.Scanner; public class SmallerThanX { public static void main(String[]args){ Scanner ab = new Scanner(System.in); int A = ab.nextInt(); int X = ab.nextInt(); int size[] = new int[A]; for(int i = 0; i<A ; i++){ int a = ab.nextInt(); size[i] = a; } for(int k = 0; k< size.length; k++){ if(X> size[k]){ System.out.print(size[k]+" "); } } } }