130.被围绕的区域.java

/*
 * @lc app=leetcode.cn id=130 lang=java
 *
 * [130] 被围绕的区域
 *
 * https://leetcode-cn.com/problems/surrounded-regions/description/
 *
 * algorithms
 * Medium (37.22%)
 * Likes:    104
 * Dislikes: 0
 * Total Accepted:    10.8K
 * Total Submissions: 28.4K
 * Testcase Example:  '[["X","X","X","X"],["X","O","O","X"],["X","X","O","X"],["X","O","X","X"]]'
 *
 * 给定一个二维的矩阵，包含 'X' 和 'O'（字母 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。
 * 
 * 示例:
 * 
 * X X X X
 * X O O X
 * X X O X
 * X O X X
 * 
 * 
 * 运行你的函数后，矩阵变为：
 * 
 * X X X X
 * X X X X
 * X X X X
 * X O X X
 * 
 * 
 * 解释:
 * 
 * 被围绕的区间不会存在于边界上，换句话说，任何边界上的 'O' 都不会被填充为 'X'。 任何不在边界上，或不与边界上的 'O' 相连的 'O'
 * 最终都会被填充为 'X'。如果两个元素在水平或垂直方向相邻，则称它们是“相连”的。
 * 
 */
class Solution {
    public void solve(char[][] board) {
        int row = board.length;
        if (row <= 1 || board[0].length <= 1) {
            return;
        }
        int col = board[0].length;

        for (int j = 0; j < col; j++) {
            solve(board, 0, j);
            solve(board, row - 1, j);
        }

        for (int i = 0; i < row; i++) {
            solve(board, i, 0);
            solve(board, i, col - 1);
        }

        for (int i = 0; i < row; i++) {
            for (int j = 0; j < col; j++) {
                if (board[i][j] == '#') {
                    board[i][j] = 'O';
                } else if (board[i][j] == 'O') {
                    board[i][j] = 'X';
                }
            }
        }
    }

    private void solve(char[][] board, int i, int j) {
        if (i == -1 || j == -1 || board.length == i || board[0].length == j) {
            return;
        }
        if (board[i][j] == 'O') {
            board[i][j] = '#';
            solve(board, i + 1, j);
            solve(board, i - 1, j);
            solve(board, i, j + 1);
            solve(board, i, j - 1);
        }
    }
}