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
思路是从外围是’O'的开始深度往里走，这时候里面的‘O'就有两种命运，一种是跟外围的’O'是联通的，那么这些‘O'就可以存活，剩下的孤立的’O'就没办法了，就只能被‘X'了，为了区分这两种’O'，我们把联通 的‘O'改为另外一种统一而独立的字符，便于后期做恢复。这就是三步走（或两步）。
1）首先从外围的‘O'处深度搜索，见到链接的’O'就把他们都改为其他标识。
2）将剩余的孤立的’O'改为‘X'，同时，将遇到标识符改为’O'。
简单吧，简单的一塌糊涂，但是这种思路太精髓了。
深度周游DFS, 大数时会因为递归太深而超时，BFS就不会超时。。
1 class Solution { 2 public: 3 void dfs(vector<vector<char> > &board, int i, int j) 4 { 5 size_t row = board.size(); 6 size_t col = board[0].size(); 7 8 if(i < 0 || i > row-1 || j < 0 || j > col-1) 9 return; 10 if(board[i][j] != 'O') 11 return; 12 board[i][j] = '*';//tag, in order to reverse back 13 dfs(board, i, j-1); 14 dfs(board, i, j+1); 15 dfs(board, i-1, j); 16 dfs(board, i+1, j); 17 } 18 19 void solve(vector<vector<char> > &board) 20 { 21 size_t row = board.size(); 22 size_t col = board[0].size(); 23 24 // dfs traverse form the up and down 25 for(int j = 0; j < col; j++) 26 { 27 dfs(board, 0 ,j); 28 dfs(board,row-1, j); 29 } 30 31 // dfs traverse form the left and right 32 for(int i = 0; i < row; i++) 33 { 34 dfs(board, i, 0); 35 dfs(board, i, col-1); 36 } 37 38 for(int i = 0; i < row; i++) 39 { 40 for(int j = 0; j < col; j++) 41 { 42 if(board[i][j] == 'O') 43 board[i][j] = 'X'; 44 else if(board[i][j] == '*') 45 board[i][j] = 'O'; 46 } 47 } 48 49 50 } 51 };
广度周游BFS
1 class Solution { 2 queue<pair<int, int> > m_que; 3 public: 4 void dfs(vector<vector<char> > &board, int i, int j) 5 { 6 size_t row = board.size(); 7 size_t col = board[0].size(); 8 9 if(i < 0 || i > row-1 || j < 0 || j > col-1) 10 return; 11 if(board[i][j] != 'O') 12 return; 13 board[i][j] = '*';//tag, in order to reverse back 14 dfs(board, i, j-1); 15 dfs(board, i, j+1); 16 dfs(board, i-1, j); 17 dfs(board, i+1, j); 18 } 19 20 void fill(vector<vector<char> > &board, int x, int y){ 21 size_t row = board.size(); 22 size_t col = board[0].size(); 23 24 if(x<0 || x>=row || y<0 || y>=col || board[x][y]!='O') 25 return; 26 27 pair<int, int> p ; 28 p.first = x; 29 p.second = y; 30 m_que.push(p); 31 32 board[x][y]='*'; 33 } 34 35 36 void bfs(vector<vector<char> > &board, int i, int j) 37 { 38 fill(board, i, j); 39 40 while(!m_que.empty()) 41 { 42 pair<int, int> p = m_que.front() ; 43 m_que.pop(); 44 45 i = p.first; 46 j = p.second; 47 48 fill(board, i, j-1); 49 fill(board, i, j+1); 50 fill(board, i-1, j); 51 fill(board, i+1, j); 52 } 53 54 } 55 56 void solve(vector<vector<char> > &board) 57 { 58 if (board.empty()) return; 59 size_t row = board.size(); 60 size_t col = board[0].size(); 61 62 // dfs traverse form the up and down 63 for(int j = 0; j < col; j++) 64 { 65 #if 0 66 dfs(board, 0 ,j); 67 dfs(board,row-1, j); 68 #endif 69 bfs(board, 0 ,j); 70 bfs(board,row-1, j); 71 72 } 73 74 // dfs traverse form the left and right 75 for(int i = 0; i < row; i++) 76 { 77 #if 0 78 dfs(board, i, 0); 79 dfs(board, i, col-1); 80 #endif 81 bfs(board, i, 0); 82 bfs(board, i, col-1); 83 84 } 85 86 for(int i = 0; i < row; i++) 87 { 88 for(int j = 0; j < col; j++) 89 { 90 if(board[i][j] == 'O') 91 board[i][j] = 'X'; 92 else if(board[i][j] == '*') 93 board[i][j] = 'O'; 94 } 95 } 96 97 98 } 99 };