DFS 02

leetcode-cn.com/problems/su…

解题思路

由于边界上的'O'不属于被包围，所以可以从边界上的'O'出发，寻找与之相连的其他'O',这一连串的 O 都是不被包围的，暂时标记为'A'，DFS完后，可以遍历整个矩阵，然后找到剩下的'O'，改成'X',再把'A'改回'O'。

代码如下

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';
                }
            }
        }
//		System.out.println(Arrays.deepToString(board));
	}

	public void dfs(char[][] board, int i, int j) {
		if (i < 0 || i >= n || j < 0 || j >= m || board[i][j] != 'O') {
			return;
		}
		board[i][j] = 'A'; // 把与边界'O'相连的所有'O'都标记成'A'

		dfs(board, i + 1, j);
		dfs(board, i - 1, j);
		dfs(board, i, j + 1);
		dfs(board, i, j - 1);
	}

//	public static void main(String[] args) {
//		char[][] a = new char[][] { { 'X', 'X', 'X', 'X' }, { 'X', 'O', 'O', 'X' }, { 'X', 'X', 'O', 'X' },
//				{ 'X', 'O', 'X', 'X' } };
//		Solution k = new Solution();
//		System.out.println(Arrays.deepToString(a));
//		k.solve(a);
//	}

}