Skip to main content

DEPTH FIRST SEARCH IN JAVA

DEPTH FIRST SEARCH IN JAVA



public void dfs()
 {
  
  vertex[0].visited=true;
  displyVertex(0);
  stackobj.push(0);
  
  while(!stackobj.isEmpty())
  {
   int v=adjVertex((Integer) stackobj.peek());
   if(v==-1)
   {
    stackobj.pop();
   }
    else{
     vertex[v].visited=true;
     displyVertex(v);
     stackobj.push(v);
     
    }
   }
  }
  

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 75 https://leetcode.com/list/5hkn6wze/ Must do 60  https://leetcode.com/list/5eie1aqd/ Must do medium:-  https://leetcode.com/list/5xaelz7g/ Must do Easy:-   https://leetcode.com/list/5r7rxpr1/ Graph:-  https://leetcode.com/list/x18ervrd/  Dynamic Programming:-    https://leetcode.com/list/x14z0dxr/  FaceBook interviews:- https://leetcode.com/list/xyu98pv6/  Amazon Interviews:-  https://leetcode.com/list/5hkniyf7/  Google Interviews:- https://leetcode.com/list/xyu9xfo1/ https://github.com/nazarmubeen/TopProblems/blob/master/README.md

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; } ...