130. Surrounded Regions

Posted 2022-11-11 wentiliangkaihua

Given 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.

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

Explanation:

Surrounded regions shouldn’t be on the border, which means that any ‘O‘ on the border of the board are not flipped to ‘X‘. Any ‘O‘ that is not on the border and it is not connected to an ‘O‘ on the border will be flipped to ‘X‘. Two cells are connected if they are adjacent cells connected horizontally or vertically.

class Solution 
    public void solve(char[][] board) 
        if (board.length == 0 || board[0].length == 0) return;
        if (board.length < 3 || board[0].length < 3) return;
        int m = board.length;
        int n = board[0].length;
        for (int i = 0; i < m; i++) 
            if (board[i][0] == ‘O‘) helper(board, i, 0);
            if (board[i][n - 1] == ‘O‘) helper(board, i, n - 1);
        
        for (int j = 1; j < n - 1; j++) 
            if (board[0][j] == ‘O‘) helper(board, 0, j);
            if (board[m - 1][j] == ‘O‘) helper(board, m - 1, j);
        
        for (int i = 0; i < m; i++) 
            for (int j = 0; j < n; j++) 
                if (board[i][j] == ‘O‘) board[i][j] = ‘X‘;
                if (board[i][j] == ‘*‘) board[i][j] = ‘O‘;
            
        
    
    
    private void helper(char[][] board, int r, int c) 
        if (r < 0 || c < 0 || r > board.length - 1 || c > board[0].length - 1 || board[r][c] != ‘O‘) return;
        board[r][c] = ‘*‘;
        helper(board, r + 1, c);
        helper(board, r - 1, c);
        helper(board, r, c + 1);
        helper(board, r, c - 1);
    

先把周围的o找到替换成*，然后把跟边界o相连的o也替换成*，然后把仍存在的O换成X，再把*换回O