Skip to main content

BREADTH FIRST SEARCH IN JAVA

BREADTH FIRST SEARCH IN JAVA:-


package com.problems.graph;

import java.awt.DisplayMode;
import java.util.Iterator;
import java.util.LinkedList;




public class BFSGraph {

 int maxsize;
 Vertex vertexlist[];
 int matrixlist[][];
 int vertexcount;
 @SuppressWarnings({ "rawtypes", "unused" })
 LinkedList queue;
 
 public BFSGraph()
 {
  maxsize=20;
  matrixlist=new int[maxsize][maxsize];
  vertexlist=new Vertex[maxsize];
  for(int i=0;i<maxsize;i++)
  {
   for(int j=0;j<maxsize;j++)
   {
    matrixlist[i][j]=0;
   }
  }
  queue= new LinkedList();
  
 }
 
 public void addVertex(char label)
 {
  vertexlist[vertexcount++]=new Vertex(label);
 }
 public void addEdge(int i,int j)
 {
  matrixlist[i][j]=1;
  matrixlist[j][i]=1;
 }
 
 public void displayVertex(int v)
 {
  System.out.println(vertexlist[v].label);
 }
 
 public int adjVertex(int v)
 {
  for(int i=0;i<maxsize;i++)
  {
   if(matrixlist[v][i]==1 && vertexlist[i].visited==false )
    return i;
  }
  return -1;
 }
 
 public void bfs()
 {
  System.out.println("in bfs");
  vertexlist[0].visited=true;
  displayVertex(0);
  queue.add(0);
  int v2;
  while(!queue.isEmpty())
  {
   int v1=(Integer) queue.remove();
   System.out.println("removed"+v1);
   while( (v2=adjVertex(v1))!=-1)
   {
   vertexlist[v2].visited=true;
   displayVertex(v2);
   queue.add(v2);
   }
  }
  for(int j=0;j<vertexcount;j++)
  {
   vertexlist[j].visited=false;
  }
 }
}

Comments

.

Popular posts from this blog

Basics of System Design

This article is first one from the series of articles dedicated to system design interviews. Here i am going to present the base scenario to consider before starting to solve system design problems. Questions to ask? 1) what is the number of requests a website will recieve in a day/month/second? 2) what is the amount of memory a website will deal in a day/month/second? 3) what is the number of servers that can accomodate these requests? To answer this , first we need to remember the below numbers:- 1 million = 10 lakh = 1000000 = 10^6 1 billion = 1000 million = 10^9 1 KB = 1024 B = 10^3 1 MB= 10^6 = 1024 KB 1 GB= 10^9 = 1024 MB 1 TB = 10^12 = 1024 GB Memory we need to see in Bytes Requests we need to see in numbers example :- suppose a website recieves 100M requests every month then:- requests per day = request per month /24 = 416700 requests requests per second = requests per day / (24*360...

Delete node at a given position in Linked List

Delete node at a given position in Linked List //delete node by position in linked list public Node deleteKeyAtPosition ( Node head , int position ) { Node temp = head ; Node prevtemp = temp ; int c = 1 ; //if position is head if ( position == 1 ) { head = head . next ; return head ; } //position +1 because we have to go till that point while ( c != position + 1 ) { if ( c == position && position != 1 ) { prevtemp . next = temp . next ; temp . next = null ; temp = prevtemp ; } prevtemp = temp ; temp = temp . next ; c ++; } return head ; }