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.

思路：

DFS。

先找到一个等于word[0]的坐标作为开始点，然后以这个点开始往四周开始查找。用visited数组记录是否访问过某个点，避免循环访问。

代码：

1     bool exist(int x, int y, string word, int index, vector
     
      
        > &visited, vector
       
        
          > &board){ 2         if(x < 0 || y < 0 || x >= board.size() || y >= board[0].size()) 3             return false; 4         if(board[x][y] != word[index] || visited[x][y]) 5             return false; 6         if(index == word.length()-1) 7             return true; 8         visited[x][y] = true; 9         bool result = exist(x-1, y, word, index+1, visited, board) 10                    || exist(x+1, y, word, index+1, visited, board)11                    || exist(x, y-1, word, index+1, visited, board)12                    || exist(x, y+1, word, index+1, visited, board);13         visited[x][y] = false;14         return result;15     }16     bool exist(vector
         
          
            > &board, string word) {17 // IMPORTANT: Please reset any member data you declared, as18 // the same Solution instance will be reused for each test case.19 int l = word.length();20 if(l == 0)21 return true;22 int m = board.size();23 if(m == 0)24 return false;25 int n = board[0].size();26 if(n == 0)27 return false;28 vector
           
            
              > visited(m, vector
             
              (n, false));29 for(int i = 0; i < m; i++){30 for(int j = 0; j < n; j++){31 if(board[i][j] == word[0] && exist(i, j, word, 0, visited, board)){32 return true;33 }34 }35 }36 return false;37 }