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

Best LeetCode Lists for Interviews

Here is a list of some of the best questions asked in interviews Must do Top interview questions Top 100 liked Must do 75  Must do 60  Must do medium Data structures Tree Graph  Dynamic Programming Company Interviews FaceBook interviews Amazon Interviews Google Interviews Github master List

EDIT DISTANCE PROBLEM - LEETCODE 72

  Given two words  word1  and  word2 , find the minimum number of operations required to convert  word1  to  word2 . You have the following 3 operations permitted on a word: Insert a character Delete a character Replace a character Example 1: Input: word1 = "horse", word2 = "ros" Output: 3 Explanation: horse -> rorse (replace 'h' with 'r') rorse -> rose (remove 'r') rose -> ros (remove 'e') Solution:- class Solution { public int minDistance(String word1, String word2) { if(word1.length()==0){ return word2.length(); } if(word2.length()==0){ return word1.length(); } if(word1.length()==0 && word2.length()==0){ return 0; } int[][] result=new int[word1.length()+1][word2.length()+1]; for(int i=0;i<=word1.length();i++){ result[i][0]=i; } ...