力扣 题目79- 单词搜索
题目
题解
回溯算法 找到开头然后对上下左右四个方向 回溯 已经遍历过的至* 直到长度与word一致
代码
1 #include<iostream> 2 #include<vector> 3 #include<unordered_map> 4 using namespace std; 5 //1是上 2是下 3是左 4是右 6 bool loop(vector<vector<char>>& board, string& word, int x, int y, int cnt) { 7 if (y < 0 || x < 0 || x >= board.size() || y >= board[0].size() || board[x][y] != word[cnt]) { 8 return 0; 9 } 10 cnt += 1; 11 if (cnt == word.size()) { 12 return 1; 13 } 14 char keep = board[x][y]; 15 board[x][y] = '*'; 16 //上 17 if (loop(board, word, x - 1, y, cnt)) { 18 return 1; 19 } 20 //下 21 if (loop(board, word, x + 1, y, cnt)) { 22 return 1; 23 } 24 //左 25 if (loop(board, word, x, y - 1, cnt)) { 26 return 1; 27 } 28 //右 29 if (loop(board, word, x, y + 1, cnt)) { 30 return 1; 31 } 32 board[x][y] = keep; 33 return 0; 34 } 35 //只看右和下 36 class Solution { 37 public: 38 bool exist(vector<vector<char>>& board, string word) { 39 //如果word长度比board大直接返回0 40 if (board.size() * board[0].size() < word.size()) { 41 return 0; 42 } 43 //找到开头进行回溯 44 for (int i = 0; i < board.size(); i++) { 45 for (int j = 0; j < board[i].size(); j++) { 46 if (board[i][j] == word[0]) { 47 int x = i; 48 int y = j; 49 int succes = loop(board, word, x, y, 0); 50 if (succes) { 51 return true; 52 } 53 } 54 } 55 } 56 return 0; 57 } 58 }; 59 int main() { 60 Solution sol; 61 vector<vector<char>> board = { {'A', 'B', 'C', 'E'},{'S', 'F', 'C', 'S' },{'A', 'D', 'E', 'E'}}; 62 string word = "ABCCED"; 63 bool result=sol.exist(board, word); 64 cout << result << endl; 65 }
