Word Search

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.

ps: 首先找到首字符，然后四个方向搜索

class Solution {private:    int flag[100][100];public:   void dfs( vector<vector<char> > &board, int x, int y, string word, bool flag[100][100], int k, bool & end_flag ){if( k == word.size()-1 && board[x][y] == word[k] ){end_flag = true;  return ;}if( end_flag ) return;//防止在找到该完整字符串时，继续向其他方向搜索。。。if( board[x][y] == word[k] ){flag[x][y] = true;if( (x-1) >= 0  && !flag[x-1][y]  && (k+1) < word.size() )dfs( board, x-1, y, word, flag, k+1, end_flag );if( (y-1) >= 0 && !flag[x][y-1] && (k+1) < word.size() )dfs( board, x, y-1, word, flag, k+1, end_flag );if( (x+1)< board.size() && !flag[x+1][y] && (k+1) < word.size() )dfs( board, x+1, y, word, flag, k+1, end_flag );if( (y+1) < board[x].size() && !flag[x][y+1]  && (k+1) < word.size() )dfs( board, x, y+1, word, flag, k+1, end_flag );flag[x][y] = false;}else return;}    bool exist(vector<vector<char> > &board, string word)     {if( !word.length() ) return true;        int col = board.size();        for( int i = 0; i < board.size(); ++i )for( int j = 0; j < board[i].size(); ++j ){if( board[i][j] == word[0] ){memset(flag,false,sizeof(flag));int k = 0;bool _end = false;dfs( board, i, j, word, flag, 0, _end );if( _end ) return true;}}return false;}};