floodfill算法系列一＞被围绕的区域

目录

• 整体思想：
• 代码设计：
• 代码呈现：

整体思想：

代码设计：

代码呈现：

class Solution {
    int m,n;
    int[] dx = {0,0,-1,1};
    int[] dy = {-1,1,0,0};

    public void solve(char[][] board) {
        m = board.length;
        n = board[0].length;

        //先扫描边界，把不满足题意的连通块变为'.'
        for(int i = 0; i < n; i++){
            if(board[0][i] == 'O') dfs(board,0,i);
            if(board[m-1][i] == 'O') dfs(board,m-1,i);
        }

        for(int j = 0; j < m; j++){
            if(board[j][0] == 'O') dfs(board,j,0);
            if(board[j][n-1] == 'O') dfs(board,j,n-1);
        }


        //最后把'.'改会'o',把'o'改为'X'
        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] == '.')  
                    board[i][j] = 'O';
            }
    }

    private void dfs(char[][] board, int i, int j){
        board[i][j] = '.';

        for(int k = 0; k < 4; k++){
            int x = dx[k] + i; int y = dy[k] + j;
            if(x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'O'){
                dfs(board,x,y);
            }
        }
    }
}