Word Search
问题描述：
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.
测试代码（c++）：
class Solution {
bool panduan(vector<vector<char>>& board,string word,int row,int column,int len)
{
if(len==word.length())
return true;
bool flag = false;
if(!flag&&row>0&&board[row-1][column]!='.'&&board[row-1][column]==word[len])
{
board[row-1][column] = '.';
flag = panduan(board,word,row-1,column,len+1);
board[row-1][column] = word[len];
}
if(!flag&&row<board.size()-1&&board[row+1][column]!='.'&&board[row+1][column]==word[len])
{
board[row+1][column] = '.';
flag = panduan(board,word,row+1,column,len+1);
board[row+1][column] = word[len];
}
if(!flag&&column<board[0].size()-1&&board[row][column+1]!='.'&&board[row][column+1]==word[len])
{
board[row][column+1] = '.';
flag = panduan(board,word,row,column+1,len+1);
board[row][column+1] = word[len];
}
if(!flag&&column>0&&board[row][column-1]!='.'&&board[row][column-1]==word[len])
{
board[row][column-1] = '.';
flag = panduan(board,word,row,column-1,len+1);
board[row][column-1] = word[len];
}
return flag;
}
public:
bool exist(vector<vector<char>>& board, string word) {
if(board.empty())
return false;
for(int i=0;i<board.size();i++)
{
for(int j=0;j<board[0].size();j++)
{
if(board[i][j]==word[0])
{
board[i][j] = '.';
if(panduan(board,word,i,j,1))
return true;
board[i][j] = word[0];
}
}
}
return false;
}
};