leetcode -- Surrounded Regions
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
[解题思路]
board的四个边上的O是无法被X包围的，故如果在四个边上发现O，则这些O应当被保留 从这些O出发继续寻找相邻的O，这些O也是要保留的
等这些O都标记结束，则剩余的O就应该被改成X
1.使用DFS解，在过大数据集时会挂掉，200*200的矩阵会栈溢出，
1 public class Solution { 2 public static void solve(char[][] board) { 3 if (board == null || board.length == 0) { 4 return; 5 } 6 int row = board.length; 7 int col = board[0].length; 8 9 // up row 10 for (int j = 0; j < col; j++) { 11 if (board[0][j] == 'O') { 12 dfs(board, 0, j); 13 } 14 } 15 16 // bottom row 17 for (int j = 0; j < col; j++) { 18 if (board[row - 1][j] == 'O') { 19 dfs(board, row - 1, j); 20 } 21 } 22 23 // left column 24 for (int i = 0; i < row; i++) { 25 if (board[i][0] == 'O') { 26 dfs(board, i, 0); 27 } 28 } 29 30 // right column 31 for (int i = 0; i < row; i++) { 32 if (board[i][col - 1] == 'O') { 33 dfs(board, i, col - 1); 34 } 35 } 36 37 for (int i = 0; i < row; i++) { 38 for (int j = 0; j < col; j++) { 39 if (board[i][j] == 'O') { 40 board[i][j] = 'X'; 41 } 42 if (board[i][j] == 'P') { 43 board[i][j] = 'O'; 44 } 45 } 46 } 47 } 48 49 private static void dfs(char[][] board, int i, int j) { 50 if(i < 0 || i >= board.length || j < 0 || j > board[0].length || board[i][j] != 'O'){ 51 return; 52 } 53 board[i][j] = 'P'; 54 55 // up 56 if (i - 1 >= 0 && board[i - 1][j] == 'O') { 57 dfs(board, i - 1, j); 58 } 59 // bottom 60 if (i + 1 < board.length && board[i + 1][j] == 'O') { 61 dfs(board, i + 1, j); 62 } 63 // left 64 if (j - 1 >= 0 && board[i][j - 1] == 'O') { 65 dfs(board, i, j - 1); 66 } 67 // right 68 if (j + 1 < board[0].length && board[i][j + 1] == 'O') { 69 dfs(board, i, j + 1); 70 } 71 72 } 73 }
2.BFS
使用BFS来对board进行遍历，如果当前格是'O'，则将该格位置放入到queue中
遍历queue，queue中当前格是'O'时，将其内容改为'P'，则将该格的上下左右都放入到queue中，
1 public class Solution { 2 private Queue<Integer> queue = new LinkedList<Integer>(); 3 public void solve(char[][] board) { 4 if (board.length == 0) { 5 return; 6 } 7 if (board[0].length == 0) { 8 return; 9 } 10 int row = board.length; 11 int col = board[0].length; 12 13 // up row 14 for (int j = 0; j < col; j++) { 15 if (board[0][j] == 'O') { 16 bfs(board, 0, j); 17 } 18 } 19 20 // bottom row 21 for (int j = 0; j < col; j++) { 22 if (board[row - 1][j] == 'O') { 23 bfs(board, row - 1, j); 24 } 25 } 26 27 // left column 28 for (int i = 0; i < row; i++) { 29 if (board[i][0] == 'O') { 30 bfs(board, i, 0); 31 } 32 } 33 34 // right column 35 for (int i = 0; i < row; i++) { 36 if (board[i][col - 1] == 'O') { 37 bfs(board, i, col - 1); 38 } 39 } 40 41 for (int i = 0; i < row; i++) { 42 for (int j = 0; j < col; j++) { 43 if (board[i][j] == 'O') { 44 board[i][j] = 'X'; 45 } 46 if (board[i][j] == 'P') { 47 board[i][j] = 'O'; 48 } 49 } 50 } 51 52 } 53 54 private void fill(char[][] board, int i, int j) { 55 int row = board.length; 56 int col = board[0].length; 57 if (i < 0 || i >= row || j < 0 || j >= col || board[i][j] != 'O') 58 return; 59 60 queue.offer(i * col + j); 61 board[i][j] = 'P'; 62 } 63 64 private void bfs(char[][] board, int i, int j) { 65 int col = board[0].length; 66 67 fill(board, i, j); 68 69 while (!queue.isEmpty()) { 70 int cur = queue.poll(); 71 int x = cur / col; 72 int y = cur % col; 73 74 fill(board, x - 1, y); 75 fill(board, x + 1, y); 76 fill(board, x, y - 1); 77 fill(board, x, y + 1); 78 } 79 } 80 }
ref:
