Word Search

Given a 2D board and a word, find if the word exists in the grid.

The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once.

For example,
Given board =

[   ["ABCE"],   ["SFCS"],   ["ADEE"] ] 

word  =  "ABCCED" , -> returns  true ,
word  =  "SEE" , -> returns  true ,

word = "ABCB", -> returns false.

class Solution {
private:
     int m,n;
     int len;
     int dep;
     char** board;
     string word;
     int x,y;
     bool search()
    {
         if(dep==len)  return  true;
         int xnew[ 4];
         int ynew[ 4];
        xnew[ 0]=x;xnew[ 1]=x;xnew[ 2]=x- 1;xnew[ 3]=x+ 1;
        ynew[ 0]=y- 1;ynew[ 1]=y+ 1;ynew[ 2]=y;ynew[ 3]=y;
         for( int i= 0;i< 4;i++)
         if(xnew[i]>= 0 && xnew[i]<m && ynew[i]>= 0 && ynew[i]<n && board[xnew[i]][ynew[i]]==word[dep])
        {
            board[xnew[i]][ynew[i]]= 0;
            dep++;
            x=xnew[i];y=ynew[i];
             if(search()) return  true;
            dep--;
            board[xnew[i]][ynew[i]]=word[dep];
        }
         return  false;
    }
public:
     bool exist(vector<vector< char> > &board,  string word) 
    {
         this->word=word;
        m=board.size();
         if(m== 0)  return  false;
        n=board[ 0].size();
         this->board= new  char*[m];
         for( int i= 0;i<m;i++)
        {
             this->board[i]= new  char[n];
             for( int j= 0;j<n;j++)
                 this->board[i][j]=board[i][j];
        }
        
        len=word.length();
         for( int i= 0;i<m;i++)
             for( int j= 0;j<n;j++)
             if( this->board[i][j]==word[ 0])
            {
                 this->board[i][j]= 0;
                x=i;y=j;dep= 1;
                 if(search())     return  true;
                 this->board[i][j]=word[ 0];
            }
         return  false;
    }
};