Day67 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

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

示例1：

image

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'

Java解法

思路：

• 初步设想，找到每个O,然后判断是否被包围，导致有各种牵扯，换个角度想想，我们只需要从O可以留下来思考便很简单了，
• 遍历最外层，判断是否有O，然后以该O为起点进行遍历标记能到达的O

package sj.shimmer.algorithm.m4_2021;

/**
 * Created by SJ on 2021/4/4.
 */

class D67 {
    public static void main(String[] args) {
        char[][] chars = new char[][]{
                {'X', 'X', 'X', 'X'},
                {'X', 'O', 'O', 'X'},
                {'X', 'X', 'O', 'X'},
                {'X', 'O', 'X', 'X'}
        };
        chars = new char[][]{{'X'}};
        chars = new char[][]{{'O','O'},{'O','O'}};
        solve(chars);
        for (char[] aChar : chars) {

            System.out.println(aChar);
        }
    }

    public static void solve(char[][] board) {
        if (board != null && board.length != 0) {
            int height = board.length;
            int width = board[0].length;
            for (int w = 0; w < width; w++) {
                //上  w<width-index,h=index-1
                if (board[0][w] == 'O') {
                    board[0][w] = 'A';
                    checkCanArriveO(board, height, width, 0, w);
                }
                //下  w<width-index,h=height-index
                if (board[height - 1][w] == 'O') {
                    board[height - 1][w] = 'A';
                    checkCanArriveO(board, height, width, height - 1, w);
                }
            }
            for (int h = 1; h < height-1; h++) {
                //左  w=index-1,h<height-index
                if (board[h][0] == 'O') {
                    board[h][0] = 'A';
                    checkCanArriveO(board, height, width, h, 0);
                }
                //右  w=w-index,h<height-index
                if (board[h][width - 1] == 'O') {
                    board[h][width - 1] = 'A';
                    checkCanArriveO(board, height, width, h, width - 1);
                }
            }
            for (int i = 0; i < height; i++) {
                for (int j = 0; j < width; j++) {
                    if (board[i][j]=='O') {
                        board[i][j]='X';
                    }
                    if (board[i][j]=='A') {
                        board[i][j]='O';
                    }
                }
            }
        }
    }

    public static void checkCanArriveO(char[][] board, int height, int width, int h, int w) {
        //上
        if (h - 1 >= 0 && board[h - 1][w] == 'O') {
            board[h - 1][w] = 'A';
            checkCanArriveO(board, height, width, h - 1, w);
        }
        //下
        if (h + 1 < height && board[h + 1][w] == 'O') {
            board[h + 1][w] = 'A';
            checkCanArriveO(board, height, width, h + 1, w);
        }
        //左
        if (w - 1 >= 0 && board[h][w - 1] == 'O') {
            board[h][w - 1] = 'A';
            checkCanArriveO(board, height, width, h, w - 1);
        }
        //右
        if (w + 1 < width && board[h][w + 1] == 'O') {
            board[h][w + 1] = 'A';
            checkCanArriveO(board, height, width, h, w + 1);
        }
    }
}

image

官方解

https://leetcode-cn.com/problems/surrounded-regions/solution/bei-wei-rao-de-qu-yu-by-leetcode-solution/

1. 深度优先搜索

   我的参考解法，但官方解写的更加优雅

   class Solution {
       int n, m;
   
       public void solve(char[][] board) {
           n = board.length;
           if (n == 0) {
               return;
           }
           m = board[0].length;
           for (int i = 0; i < n; i++) {
               dfs(board, i, 0);
               dfs(board, i, m - 1);
           }
           for (int i = 1; i < m - 1; i++) {
               dfs(board, 0, i);
               dfs(board, n - 1, i);
           }
           for (int i = 0; i < n; i++) {
               for (int j = 0; j < m; j++) {
                   if (board[i][j] == 'A') {
                       board[i][j] = 'O';
                   } else if (board[i][j] == 'O') {
                       board[i][j] = 'X';
                   }
               }
           }
       }
   
       public void dfs(char[][] board, int x, int y) {
           if (x < 0 || x >= n || y < 0 || y >= m || 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);
       }
   }
   

   • 时间复杂度：O(n×m)
   • 空间复杂度：O(n×m)

2. 广度优先搜索

   这是我最开始遍历外圈到内的解法正解，搞定了我无法对存储标记点的存储方案（使用list记录长度为2的int[]数组），最后标记遍历。