#leetcode#130. Surrounded Regions

iven a 2D board containing 'X' and 'O' (the letter 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 XX O O XX X O XX O X X

After running your function, the board should be:

X X X XX X X XX X X XX O X X

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思路是flood fill，先把四条边上的O染色，把相邻O变成‘1’， 然后再遍历一般即可

class Solution {    public void solve(char[][] board) {        if(board == null || board.length < 2  || board[0].length < 2)            return;        int m = board.length;         int n = board[0].length;                for(int i = 0; i < n; i++){            fillEdge(board, 0, i);            fillEdge(board, m - 1, i);        }        for(int j = 0; j < m; j++){            fillEdge(board, j, 0);            fillEdge(board, j, n - 1);        }                        for(int i = 0; i < m; i++){            for(int j = 0; j < n; j++){                if(board[i][j] == 'O'){                    board[i][j] = 'X';                }else if(board[i][j] == '1'){                    board[i][j] = 'O';                }            }        }    }        private void fillEdge(char[][] board, int row, int col){        if(board[row][col] != 'O'){            return;        }        int m = board.length;        int n = board[0].length;        LinkedList<Integer> queue = new LinkedList<>();        queue.offer(row * n + col);        while(!queue.isEmpty()){            int idx = queue.poll();            int x = idx / n;            int y = idx % n;            if(x < 0 || x >= m || y < 0 || y >= n || board[x][y] != 'O'){                continue;            }            board[x][y] = '1';            queue.offer((x + 1) * n + y);            queue.offer((x - 1) * n + y);            queue.offer(x * n + y + 1);            queue.offer(x * n + y - 1);        }    }    }