力扣:130. 被围绕的区域(深度优先算法)

超、凢脫俗 2024-03-31 12:43 57阅读 0赞

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'

来源：力扣（LeetCode）  
链接：https://leetcode.cn/problems/surrounded-regions  
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

class Solution {
        int n;
        int m;
       public void solve(char[][] board) {
            n = board.length;
            m = board[0].length;
    
            //将边界的元素调用dfs方法.
            for (int i = 0; i <= n - 1; i++) {
                dfs(board, i, 0);
                dfs(board, i, m - 1);
            }
            for (int i = 1; i <= m - 2; i++) {
                dfs(board, 0, i);
                dfs(board, n - 1, i);
            }
    
            //遍历二维数组,若被标记为A则为边界还原为O,否则改变为X.
            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';
                    }
                }
    
            }
        }
    
    
        //标记边界及其连接的'O';
        public void dfs(char[][] board, int x, int y) {
            //递归边界,如果超过边界或者不为'O'.
            if (x < 0 || x > n - 1 || y < 0 || y > m - 1 || 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);
        }
    
        
    }