Given a 2D board containing 'X' and 'O', capture all regions surrounded by 'X'.
A region is captured by flipping all 'O's into 'X's in that surrounded region .
For example,
X X X X X O O X X X O X X O X X
After running your function, the board should be:
X X X X X X X X X X X X X O X X
思路：
类似以前写的迷宫类题目，递归实现，但是调试的过程中还是费了些功夫，头脑不够清晰。
通过观察如果O区域有一块在边缘，那么就不会变成X，其余的O区域都会变成X。所以直接从四条边上的O开始扩散，用另一个vector记录是否有扩散到，最后根据那个vector的结果重新给原board赋值即可。
1 vector<vector<bool> > v; 2 int m,n; 3 void visit(int row, int col, vector<vector<char> > &board){ 4 if(row < 0 || row > m-1 || col < 0 || col > n-1) 5 return; 6 if(board[row][col] == 'X') 7 return; 8 v[row][col] = true; 9 board[row][col] = 'X'; 10 visit(row+1, col, board); 11 visit(row-1, col, board); 12 visit(row, col+1, board); 13 visit(row, col-1, board); 14 } 15 void solve(vector<vector<char> > &board) { 16 // IMPORTANT: Please reset any member data you declared, as 17 // the same Solution instance will be reused for each test case. 18 m = board.size(); 19 if(m == 0) 20 return; 21 n = board[0].size(); 22 if(n == 0) 23 return; 24 v = vector<vector<bool> >(m,vector<bool>(n,false)); 25 int i,j; 26 for(i = 0; i < n; i++){ 27 if(board[0][i] == 'O') 28 visit(0, i, board); 29 } 30 for(i = 0; i < n; i++){ 31 if(board[m-1][i] == 'O') 32 visit(m-1, i, board); 33 } 34 for(i = 0; i < m; i++){ 35 if(board[i][0] == 'O') 36 visit(i, 0, board); 37 } 38 for(i = 0; i < m; i++){ 39 if(board[i][n-1] == 'O') 40 visit(i, n-1, board); 41 } 42 for(i = 0; i < m; i++){ 43 for(j = 0; j < n; j++){ 44 board[i][j] = v[i][j] ? 'O':'X'; 45 } 46 } 47 }
第二遍。原先的代码发现不能AC了，可能是数据变大了，栈溢出，所以改成宽搜。
1 void visit(int x, int y, vector<vector<char> > &board, vector<vector<int> > &maze){ 2 int m = board.size(), n = board[0].size(); 3 queue<pair<int, int> > Queue; 4 Queue.push(pair<int, int>(x,y)); 5 while(!Queue.empty()){ 6 pair<int, int> tmp = Queue.front(); 7 Queue.pop(); 8 int i = tmp.first; 9 int j = tmp.second; 10 if(i < 0 || j < 0 || i >= m || j >= n) 11 continue; 12 if(board[i][j] == 'X' || maze[i][j]) 13 continue; 14 maze[i][j] = 1; 15 Queue.push(pair<int, int>(i+1, j)); 16 Queue.push(pair<int, int>(i-1, j)); 17 Queue.push(pair<int, int>(i, j+1)); 18 Queue.push(pair<int, int>(i, j-1)); 19 } 20 } 21 void solve(vector<vector<char> > &board) { 22 // IMPORTANT: Please reset any member data you declared, as 23 // the same Solution instance will be reused for each test case. 24 int m = board.size(); 25 if(m == 0) 26 return; 27 int n = board[0].size(); 28 vector<vector<int> > maze(m, vector<int>(n, 0)); 29 int i,j; 30 for(i = 0; i < n; i++){ 31 if(board[0][i] == 'O' && maze[0][i] == 0) 32 visit(0, i, board, maze); 33 if(board[m-1][i] == 'O' && maze[m-1][i] == 0) 34 visit(m-1, i, board, maze); 35 } 36 for(i = 0; i < m; i++){ 37 if(board[i][0] == 'O' && maze[i][0] == 0) 38 visit(i, 0, board, maze); 39 if(board[i][n-1] == 'O' && maze[i][n-1] == 0) 40 visit(i, n-1, board, maze); 41 } 42 for(i = 0; i < m; i++){ 43 for(j = 0; j < n; j++){ 44 if(!maze[i][j]) 45 board[i][j] = 'X'; 46 } 47 } 48 }