README.md

被围绕的区域

难度：中等

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

题目

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

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"]]

解题

深度优先搜索

/**
 * 深度优先搜索
 * @desc 时间复杂度 O(MN)  空间复杂度 O(MN)
 * @param board
 */
export function solve(board: string[][]): void {
  const m = board.length;
  if (m === 0) return;
  const n = board[0].length;
  if (n === 0) return;

  // 从最外圈向内寻找
  let i;
  for (i = 0; i < m; i++) {
    dfs(board, i, 0);
    dfs(board, i, n - 1);
  }

  for (i = 1; i < n - 1; i++) {
    dfs(board, 0, i);
    dfs(board, m - 1, i);
  }

  // 将O换成X，将A换成O
  for (i = 0; i < m; i++) {
    for (let j = 0; j < n; j++) {
      if (board[i][j] === 'A') {
        board[i][j] = 'O';
      } else if (board[i][j] === 'O') {
        board[i][j] = 'X';
      }
    }
  }

  function dfs(board: string[][], x: number, y: number) {
    if (x < 0 || x >= m || y < 0 || y >= n || board[x][y] !== 'O') return;
    board[x][y] = 'A';
    dfs(board, x + 1, y);
    dfs(board, x - 1, y);
    dfs(board, x, y + 1);
    dfs(board, x, y - 1);
  }
}

广度优先搜索

/**
 * 广度优先搜索
 * @desc 时间复杂度 O(MN)  空间复杂度 O(MN)
 * @param board
 */
export function solve2(board: string[][]): void {
  const m = board.length;
  if (m === 0) return;
  const n = board[0].length;
  if (n === 0) return;

  const queue: [number, number][] = [];
  for (let i = 0; i < m; i++) {
    if (board[i][0] === 'O') {
      queue.unshift([i, 0]);
      board[i][0] = 'A';
    }

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

    if (board[m - 1][i] === 'O') {
      queue.unshift([m - 1, i]);
      board[m - 1][i] = 'A';
    }
  }

  const dirs = [
    [0, 1],
    [1, 0],
    [0, -1],
    [-1, 0]
  ];

  while (queue.length) {
    const [x, y] = queue.pop()!;
    for (const dir of dirs) {
      const mx = x + dir[0];
      const my = y + dir[1];
      if (mx < 0 || mx >= m || my < 0 || my >= n || board[mx][my] !== 'O')
        continue;

      queue.unshift([mx, my]);
      board[mx][my] = 'A';
    }
  }

  // 将O换成X，将A换成O
  for (let i = 0; i < m; i++) {
    for (let j = 0; j < n; j++) {
      if (board[i][j] === 'A') {
        board[i][j] = 'O';
      } else if (board[i][j] === 'O') {
        board[i][j] = 'X';
      }
    }
  }
}