leetcode 56: Word Search

Word SearchApr
18 '12

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

.

uncompleted

class Position{
public:
int x;
int y;
Position(int xx, int yy) : x(xx),y(yy){};

};

class Solution {
public:
bool exist(vector<vector<char> > &board, string word) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if(word.size() ==0) return true;
if(board.size()==0) return false;
Solution::m = board.size();
Solution::n = board[0].size();
vector<bool> temp(n,false);
vector<vector<bool> > flags(m, temp);

for( int i=0; i<m; i++) {
for( int j=0; j<n; j++) {
if(board[i][j]==word[0]){
flags[i][j] = true;
if( dfs( board, word, flags, 0, Position(i, j) ) ) return true;
flags[i][j] = false;
}
}
}
return false;
}

private:
int m;
int n;

bool dfs( vector<vector<char> > &board, string & word, vector<vector<bool> >& flags, int level,Position p ) {
if( level == word.size()-1) return true;
if( word[level] != board[p.x][p.y]) return false;
for( int i=-1; i<=1; i++) {
for( int j=-1; j<=1; j++) {
if(i+j!=1 && i+j!=-1) continue;
if( p.x+i>=0 && p.y+j>=0 && p.x+i<m && p.y+j<n && !flags[p.x+i][p.y+j] ) {
flags[p.x+i][p.y+j] = true;
if( dfs(board, word, flags, level+1, Position( p.x+i, p.y+j) ) )
return true;
else
flags[p.x+i][p.y+j] == false;
}
}
}
return false;
}
};