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.
思路:遍历整个网格,只要找到一组解就返回true,在每一组解的寻找过程中采用深度优先搜索来做。
class Solution { public: bool dfs(vector<vector<char>>& board, string word, int k, int i, int j, vector<vector<int>> &aa){ if (k == word.size()) return true; if (i<0 || i >= board.size() || j<0 || j >= board[0].size() || board[i][j]!=word[k]) return false; if (aa[i][j]==0){ aa[i][j] = 1;//递归前修改 if (dfs(board, word, k + 1, i + 1, j, aa) || dfs(board, word, k + 1, i - 1, j, aa) || dfs(board, word, k + 1, i, j + 1, aa) || dfs(board, word, k + 1, i, j - 1, aa)) return true; aa[i][j] = 0;//递归后恢复 } return false; } bool exist(vector<vector<char>>& board, string word) { if (board.empty() || board[0].empty() || word.empty()) return false; int m = board.size(), n = board[0].size(); vector<vector<int>> flag(m, vector<int>(n, 0)); for (int i = 0; i < m; ++i){ for (int j = 0; j < n; ++j){ if (dfs(board, word, 0, i, j, flag)) return true; } } return false; } };