[Leetcode] Word Search

题目：
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

.

bool dfs(vector<vector<char> > &board, int x, int y, string word)
{
if(word.size() == 0)return true;

bool flag = false;
if(x-1>=0 && board[x-1][y] == word[0])
{
board[x-1][y] = '#';
flag = dfs(board, x-1, y, word.substr(1));
board[x-1][y] = word[0];
}

if(!flag && y-1>=0 && board[x][y-1] == word[0])
{
board[x][y-1] = '#';
flag = dfs(board, x, y-1, word.substr(1));
board[x][y-1] = word[0];
}

if(!flag && x+1<board.size() && board[x+1][y] == word[0])
{
board[x+1][y] = '#';
flag = dfs(board, x+1, y, word.substr(1));
board[x+1][y] = word[0];
}

if(!flag && y+1<board[0].size() && board[x][y+1] == word[0])
{
board[x][y+1] = '#';
flag = dfs(board, x, y+1, word.substr(1));
board[x][y+1] = word[0];
}

return flag;
}
bool exist(vector<vector<char> > &board, string word) {
// Note: The Solution object is instantiated only once.
if(word.size() < 1)return true;
int row = board.size();
int col = board[0].size();
set<string> st;
for(int i = 0; i < row; i++)
for(int j = 0; j < col; j++)
if(board[i][j] == word[0])
{
board[i][j] = '#';
if(dfs(board,i,j,word.substr(1)))
return true;
board[i][j] = word[0];
}
return false;
}