给定一个二维网格和一个单词，找出该单词是否存在于网格中。
单词必须按照字母顺序，通过相邻的单元格内的字母构成，其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母不允许被重复使用。
示例:
board = [ ['A','B','C','E'], ['S','F','C','S'], ['A','D','E','E'] ] 给定 word = "ABCCED", 返回 true. 给定 word = "SEE", 返回 true. 给定 word = "ABCB", 返回 false.
1 #include "_000库函数.h" 2 3 //此问题是一个DFS问题，每个单词可以像上下左右四个方向进行下一步行走 4 5 6 7 class Solution { 8 public: 9 bool exist(vector<vector<char>>& board, string word) { 10 if (board.empty() || board[0].empty())return false; 11 int m = board.size(), n = board[0].size(); 12 vector<vector<bool>>visted(m, vector<bool>(n, false));//记录是否被访问过 13 for (int i = 0; i < m; ++i) { 14 for (int j = 0; j < n; ++j) { 15 if (search(board, visted, 0, i, j, word))return true;//每个单词单元都进行上下左右进行搜索！ 16 } 17 } 18 } 19 20 bool search(vector<vector<char>>& board, vector<vector<bool>>& visted, int idx, int i, int j, string word) { 21 if (idx == word.size())return true; 22 int m = board.size(), n = board[0].size(); 23 if (i < 0 || j < 0 || i >= m || j >= n || visted[i][j] || board[i][j] != word[idx])return false; 24 visted[i][j] = true; 25 bool res = (search(board, visted, idx + 1, i - 1, j, word) || 26 search(board, visted, idx + 1, i + 1, j, word) || 27 search(board, visted, idx + 1, i, j - 1, word) || 28 search(board, visted, idx + 1, i, j + 1, word)); 29 visted[i][j] = false;//回溯 30 return res; 31 } 32 }; 33 34 void T079() { 35 Solution s; 36 vector<vector<char>> board; 37 string word; 38 board = { {'A', 'B', 'C', 'E'}, {'S', 'F', 'C', 'S'}, {'A', 'D', 'E', 'E'} }; 39 word = "ABCCED"; 40 cout << s.exist(board, word) << endl; 41 42 43 }