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 =
[ ['A','B','C','E'], ['S','F','C','S'], ['A','D','E','E'] ]
word =
"ABCCED", -> returns
true,
word =
"SEE", -> returns
true,
word =
"ABCB", -> returns
false.
访问过的元素不能再访问,发现大家的实现都是用个附加结构标记访问过的.就地赋值个'\0'后面再恢复好啦.......
bool exist(vector<vector<char>>& board,int i,int j,string::iterator beg,string::iterator end)
{
bool res=true;
char cur=*beg++;
if(board[i][j]!=cur)return false;
if(beg==end)return true;
board[i][j]=0;
do{//上下左右
if(i+1<board.size()&&exist(board,i+1,j,beg,end))
break;
if(i-1>=0&&exist(board,i-1,j,beg,end))
break;
if(j+1<board[0].size()&&exist(board,i,j+1,beg,end))
break;
if(j-1>=0&& exist(board,i,j-1,beg,end))
break;
res=false;
}while(0);
board[i][j]=cur;
return res;
}
bool exist(vector<vector<char>>& board, string word) {
char beg=word[0];
for(int i=0;i<board.size();++i)
for(int j=0;j<board[0].size();++j)
if(exist(board,i,j,word.begin(),word.end()))
return true;
return false;
}