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 =
代码语言：javascript复制[
['A','B','C','E'],
['S','F','C','S'],
['A','D','E','E']
]word = "ABCCED", -> returns true,
word = "SEE", -> returns true,
word = "ABCB", -> returns false.
在矩阵中找单词。
又是一个深搜，代码不够精简，写死了。
代码语言：javascript复制class Solution {
public:
bool dfs(vector<vector<char>>& board, string word,int x,int y,vector<vector<bool>>& vis)
{
if(word.empty()) return true;
if(x 1<board.size() && word[0]==board[x 1][y] && !vis[x 1][y])
{
vis[x 1][y]=true;
if(dfs(board,word.substr(1,word.size()-1),x 1,y,vis)) return true;
vis[x 1][y]=false;
}
if(x-1>=0 && word[0]==board[x-1][y] && !vis[x-1][y])
{
vis[x-1][y]=true;
if(dfs(board,word.substr(1,word.size()-1),x-1,y,vis)) return true;
vis[x-1][y]=false;
}
if(y 1<board[0].size() && word[0]==board[x][y 1] && !vis[x][y 1])
{
vis[x][y 1]=true;
if(dfs(board,word.substr(1,word.size()-1),x,y 1,vis)) return true;
vis[x][y 1]=false;
}
if(y-1>=0 && word[0]==board[x][y-1] && !vis[x][y-1])
{
vis[x][y-1]=true;
if(dfs(board,word.substr(1,word.size()-1),x,y-1,vis)) return true;
vis[x][y-1]=false;
}
return false;
}
bool exist(vector<vector<char>>& board, string word) {
if(word.empty()) return true;
vector<bool> temp(board[0].size(),false);
vector<vector<bool>> vis(board.size(),temp);
for(int i=0;i<board.size();i )
for(int j=0;j<board[0].size();j )
if(word[0]==board[i][j])
{
vis[i][j]=true;
if(dfs(board,word.substr(1,word.size()-1),i,j,vis)) return true;
vis[i][j]=false;
}
return false;
}
};
