【LeeCode】130. 被围绕的区域

【题目描述】

给你一个 m x n 的矩阵 board ，由若干字符 'X' 和 'O' ，找到所有被 'X' 围绕的区域，并将这些区域里所有的 'O' 用 'X' 填充。

https://leetcode.cn/problems/surrounded-regions/description/

【示例】

【代码】leecode

package com.company;
import java.util.*;

// 2023-04-26
class Solution {
    public void solve(char[][] board) {
        if (board == null || board.length == 0 )return;
        int m = board.length;
        int n = board[0].length;

        for (int i = 0; i < m; i++){
            for (int j = 0; j < n; j++){
                // 这里判断的是边缘的部分
                boolean isEdge = i == 0 || j == 0 || i == m - 1 || j == n - 1;
                // 边缘存在 0
                if (isEdge && board[i][j] == 'O'){
                    dfs(board, i, 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';
                }
            }
        }
    }

    public void dfs(char[][] board, int i, int j) {
        if (i < 0 || j < 0 || i >= board.length || j >= board[0].length || board[i][j] == '#' || board[i][j] == 'X') return;
        // 说明已经搜索过了
        // 遍历后, 所有非边缘的 0 都变成了 #
        board[i][j] = '#';

        // 上 下 左 右
        dfs(board, i - 1, j);
        dfs(board, i + 1, j);
        dfs(board, i, j - 1);
        dfs(board, i, j + 1);

    }
}
public class Test {
    public static void main(String[] args) {
        // 输出：[["X","X","X","X"],["X","X","X","X"],["X","X","X","X"],["X","O","X","X"]]
        new Solution().solve(new char[][] {{'X','X','X','X'},{'X','O','O','X'},{'X','X','O','X'},{'X','O','X','X'}});
        new Solution().solve(new char[][] {{'X'}}); // 输出：['X']
    }
}