终于搞定了这个DFS，最近这个DFS写的很不顺手，我一直以为递归这种东西只是在解重构时比较麻烦，现在看来，连最简单的返回true和false的逻辑关系都不能说one hundred present 搞定。
人品啊TLE:
1 class Solution { 2 public: 3 bool legal(int i, int j, vector<vector<char>> board) 4 { 5 if (i >= 0 && i < board.size() && j >= 0 && j < board[i].size()) 6 return true; 7 return false; 8 } 9 10 bool dfs(vector<vector<char>>& board, string word, int t, int i, int j, vector<vector<bool>> visited) 11 { 12 if (t == word.length()) 13 return true; 14 if (board[i][j] == word[t]) 15 { 16 visited[i][j] = true; 17 if (legal(i - 1, j, board) && !visited[i - 1][j] && dfs(board, word, t + 1, i - 1, j, visited)) 18 return true; 19 if (legal(i, j + 1, board) && !visited[i][j + 1] && dfs(board, word, t + 1, i, j + 1, visited)) 20 return true; 21 if (legal(i + 1, j, board) && !visited[i + 1][j] && dfs(board, word, t + 1, i + 1, j, visited)) 22 return true; 23 if (legal(i, j - 1, board) && !visited[i][j - 1] && dfs(board, word, t + 1, i, j - 1, visited)) 24 return true; 25 } 26 visited[i][j] = false; 27 return false; 28 } 29 30 bool exist(vector<vector<char>>& board, string word) 31 { 32 int m = board.size(); 33 int n = board[0].size(); 34 vector<vector<bool>> visited(m, vector<bool>(n, false)); 35 for (int i = 0; i < board.size(); i++) 36 { 37 for (int j = 0; j < board[0].size(); j++) 38 { 39 if (dfs(board, word, 0, i, j, visited)) 40 return true; 41 } 42 } 43 return false; 44 } 45 };
为什么？
下面的依然是超时的， 无论怎么尝试都是超时的，觉得和提交的AC代码并没有什么区别啊？
1 #include<iostream> 2 #include<vector> 3 4 using namespace std; 5 6 7 bool isLegal(int i, int j, vector<vector<char>> board) 8 { 9 int H = board.size(); 10 int L = board[0].size(); 11 if (i >= 0 && i < H&&j >= 0 && j < L) 12 return true; 13 return false; 14 } 15 16 bool searchWord(vector<vector<char>> board, vector<vector<bool>> visited, string word, int i, int j, int index) 17 { 18 if (index >= word.length()) 19 { 20 return true; 21 } 22 /*if (word[index] == board[i][j]) 23 {*/ 24 visited[i][j] = true; 25 int dx[4] = { 1, 0, -1, 0 }; 26 int dy[4] = { 0, 1, 0, -1 }; 27 28 for (int x = 0; x < 4; ++x) 29 { 30 int ii = i + dx[x]; 31 int jj = j + dy[x]; 32 if (isLegal(ii, jj, board) && !visited[ii][jj] && board[ii][jj] == word[index] && searchWord(board, visited, word, ii, jj, index + 1)) 33 return true; 34 } 35 36 /*if (isLegal(i - 1, j, board) && !visited[i - 1][j] && board[i - 1][j]==word[index] && searchWord(board, visited, word, i - 1, j, index + 1)) 37 return true; 38 if (isLegal(i, j - 1, board) && !visited[i][j - 1] && board[i][j - 1] == word[index] && searchWord(board, visited, word, i, j - 1, index + 1)) 39 return true; 40 if (isLegal(i + 1, j, board) && !visited[i + 1][j] && board[i + 1][j] == word[index] && searchWord(board, visited, word, i + 1, j, index + 1)) 41 return true; 42 if (isLegal(i, j + 1, board) && !visited[i][j + 1] && board[i][j + 1] == word[index] && searchWord(board, visited, word, i, j + 1, index + 1)) 43 return true;*/ 44 /*}*/ 45 visited[i][j] = false; 46 return false; 47 } 48 49 bool exist(vector<vector<char>>& board, string word) { 50 if (word.empty()) return false; 51 int H = board.size(); 52 int L = board[0].size(); 53 int i, j; 54 vector<vector<bool>> visited(H, vector<bool>(L, false)); 55 for (i = 0; i < H; i++) 56 { 57 for (j = 0; j < L; j++) 58 { 59 if (word[0]==board[i][j]&&searchWord(board, visited, word, i, j, 1)) 60 { 61 return true; 62 } 63 } 64 } 65 return false; 66 } 67 68 69 int main() 70 { 71 char boardArray[][4] = 72 { 73 { 'A', 'B', 'C', 'E' }, 74 { 'S', 'F', 'C', 'S' }, 75 { 'A', 'D', 'E', 'E' } 76 }; 77 vector<vector<char>> board1(3); 78 for (int i = 0; i < 3; i++) 79 { 80 vector<char> temp(4); 81 temp[0] = 'A'; 82 board1[i] = temp; 83 for (int j = 0; j < 4; j++) 84 { 85 board1[i][j] = boardArray[i][j]; 86 } 87 } 88 89 if (exist(board1, "ABCCED")) 90 cout << 1<<endl; 91 else 92 cout << 0 << endl; 93 }