DFS-130被围绕的区域

130. 被围绕的区域

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。

示例 1：

输入： board = [["X","X","X","X"],["X","O","O","X"],["X","X","O","X"],["X","O","X","X"]]
输出： [["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。

示例 2：

输入： board = [["X"]]
输出： [["X"]]

提示：

• m == board.length
• n == board[i].length
• 1 <= m, n <= 200
• board[i][j] 为 'X' 或 'O'

/**
 * @param {character[][]} board
 * @return {void} Do not return anything, modify board in-place instead.
 */
var solve = function(board) {
  const m = board.length;
  const n = board[0].length;
  // m = 0;
  for (let i = 0; i < n; i++) {
    if (board[0][i] === 'O') {
      board[0][i] = 'A';
      dfs(m, n, board, 0, i);
    }
    if (board[m - 1][i] === 'O') {
      board[m - 1][i] = 'A';
      dfs(m, n, board, m - 1, i);
    }
  }

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

    if (board[i][n - 1] === 'O') {
      board[i][n - 1] = 'A';
      dfs(m, n, board, i, n - 1);
    }
  }

  for (let i = 0; i < m; i++) {
    for (let j = 0; j < n; j++) {
      if (board[i][j] === 'O') {
        board[i][j] = 'X';
      }

      if (board[i][j] === 'A') {
        board[i][j] = 'O';
      }
    }
  }
  return board;
};

function dfs(m, n, board, x, y) {
  if (x - 1 >= 0 && board[x - 1][y] === 'O') {
    board[x - 1][y] = 'A';
    dfs(m, n, board, x - 1, y);
  }

  if (x + 1 < m && board[x + 1][y] === 'O') {
    board[x + 1][y] = 'A';
    dfs(m, n, board, x + 1, y);
  }

  if (y - 1 >=0 && board[x][y - 1] === 'O') {
    board[x][y - 1] = 'A';
    dfs(m, n, board, x, y - 1);
  }

  if (y + 1 < n && board[x][y + 1] === 'O') {
    board[x][y + 1] = 'A';
    dfs(m, n, board, x, y + 1);
  }
}