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.
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;
}
};
