130. Surrounded Regions(M)
130.Add to List 130. Surrounded Regions
1 Given a 2D board containing 'X' and 'O' (the letter O), capture all regions surrounded by 'X'. 2 3 A region is captured by flipping all 'O's into 'X's in that surrounded region. 4 5 For example, 6 X X X X 7 X O O X 8 X X O X 9 X O X X 10 After running your function, the board should be: 11 12 X X X X 13 X X X X 14 X X X X 15 X O X X
1 /* 2 60 / 60 test cases passed. 3 Status: Accepted 4 Runtime: 13 ms 5 */ 6 7 class Solution { 8 public: 9 void dfs(vector<vector<char>>& board, int x, int y, int r, int c, vector<vector<bool>> &visited) 10 { 11 if(!visited[x][y] && board[x][y] == 'O') 12 { 13 visited[x][y] = true; 14 board[x][y] = '#'; 15 if((x > 0) && (!visited[x-1][y]) && (board[x-1][y] == 'O')) dfs(board, x-1, y, r, c, visited); // top 16 if((x+1 < r) && (!visited[x+1][y]) && (board[x+1][y] == 'O')) dfs(board, x+1, y, r, c, visited); // bottom 17 if((y > 1) && (!visited[x][y-1]) && (board[x][y-1] == 'O')) dfs(board, x, y-1, r, c, visited); // left ??? why "y > 0" can't pass the last test case. 18 if((y+1 < c) && (!visited[x][y+1]) && (board[x][y+1] == 'O')) dfs(board, x, y+1, r, c, visited); // right 19 } 20 return; 21 } 22 23 void solve(vector<vector<char>>& board) 24 { 25 size_t rlen = board.size(); 26 if(0 == rlen) return; 27 size_t clen = board[0].size(); 28 29 cout << board.size() << " " << board[0].size() << endl; 30 vector<vector<bool>> visited(rlen, vector<bool>(clen, false)); 31 32 for(int i=0; i<clen; i++) dfs(board, 0, i, rlen, clen, visited); // top 33 for(int j=0; j<clen; j++) dfs(board, rlen-1, j, rlen, clen, visited); // bottom 34 for(int m=0; m<rlen; m++) dfs(board, m, 0, rlen, clen, visited); // left 35 for(int n=0; n<rlen; n++) dfs(board, n, clen-1, rlen, clen, visited); // right 36 37 for(int i = 0; i < rlen; i++) 38 for(int j = 0; j < clen; j++) 39 { 40 if(board[i][j] == 'O') 41 board[i][j] = 'X'; 42 if(board[i][j] == '#') 43 board[i][j] = 'O'; 44 } 45 } 46 47 };
1 /* 2 Concise 12ms C++ DFS solution 3 https://discuss.leetcode.com/topic/45119/concise-12ms-c-dfs-solution 4 */ 5 class Solution { //by Kenigma 6 public: 7 void solve(vector<vector<char>>& board) { 8 if (board.empty()) return; 9 int row = board.size(), col = board[0].size(); 10 for (int i = 0; i < row; ++i) { 11 check(board, i, 0); // first column 12 check(board, i, col - 1); // last column 13 } 14 for (int j = 1; j < col - 1; ++j) { 15 check(board, 0, j); // first row 16 check(board, row - 1, j); // last row 17 } 18 for (int i = 0; i < row; ++i) 19 for (int j = 0; j < col; ++j) 20 if (board[i][j] == 'O') board[i][j] = 'X'; 21 else if (board[i][j] == '1') board[i][j] = 'O'; 22 } 23 24 void check(vector<vector<char>>& board, int i, int j) { 25 if (board[i][j] == 'O') { 26 board[i][j] = '1'; 27 if (i > 1) check(board, i - 1, j); 28 if (j > 1) check(board, i, j - 1); 29 if (i + 1 < board.size()) check(board, i + 1, j); 30 if (j + 1 < board[0].size()) check(board, i, j + 1); 31 } 32 } 33 };
