Word Search

Q：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.

A：DFS

bool dfs(vector<vector<char> > &board, string &word,int i,int j,int cur,vector<vector<bool> > &visit)
{
    if(board[i][j]!=word[cur])
		return false;
	if(cur==word.size()-1)
		return true;
	visit[i][j] = true;
	if(i+1<board.size()&&!visit[i+1][j]&&dfs(board,word,i+1,j,cur+1,visit))
		return true;
	if(i-1>=0&&!visit[i-1][j]&&dfs(board,word,i-1,j,cur+1,visit))
		return true;
	if(j+1<board[i].size()&&!visit[i][j+1]&&dfs(board,word,i,j+1,cur+1,visit))
		return true;
	if(j-1>=0&&!visit[i][j-1]&&dfs(board,word,i,j-1,cur+1,visit))
		return true;
	visit[i][j] = false;
	return false;
}

bool exist(vector<vector<char> > &board, string word) {
	// Start typing your C/C++ solution below
	// DO NOT write int main() function
	if(word.empty())
		return true;
	if(board.size()==0)
		return false;

	vector<vector<bool> >visit;

	for(int i=0;i<board.size();i++)
		visit.push_back(vector<bool>(board[i].size(),false));

	for(int i=0;i<board.size();i++)
	{
		for(int j=0;j<board[i].size();j++)
		{
			if(dfs(board,word,i,j,0,visit))
				return true;
		}
	}

	return false; 
}