130. 被围绕的区域

题目链接:130. 被围绕的区域

思路:使用两遍dfs,第一遍找到可以被替换区域的可进入点并记录,第二遍就从所有的可进入点入手遍历区域内所有点并替换。

这是我的思路,感觉还是挺新颖的(应该很少有人这样想我觉得),算是抛砖引玉了。

class Solution {
    int[] dx = {0, 0, 1, -1};
    int[] dy = {1, -1, 0, 0};
    boolean exit = false;// 判断是否可以退出
    public void solve(char[][] board) {
        // 定义一个记录矩阵判断某个点是否被走过
        int[][] record = new int[board.length][board[0].length];
        // 用一个数列存所有的可以进入点
        ArrayList<int[]> list = new ArrayList<>();
        // 遍历这个矩阵
        for(int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[0].length; j++) {
                // 如果这个点没有被走过,并且是'O',则开始深度优先搜索
                if (record[i][j] == 0 && board[i][j] == 'O') {// 某个区域的进入点
                    record[i][j] = 1;
                    exit = false;
                    dfsOnlyWalk(board, record, i, j);
                    if(!exit) {
                        list.add(new int[]{i, j});
                    }
                }
            }
        }
        for(int[] ints : list) {
            int x = ints[0], y = ints[1];
            board[x][y] = 'X';
            dfsOnlyMark(board, record, x, y);
        }
    }

    public void dfsOnlyWalk(char[][] board, int[][] record, int i, int j) {
        //区域中有点位于边缘
        if(i == 0 || i == board.length - 1 || j == 0 || j == board[0].length - 1) {
            exit = true;
        }
        for(int k = 0; k < 4; k++) {
            int xx = i + dx[k];
            int yy = j + dy[k];
            if(xx >= 0 && xx < board.length && yy >= 0 && yy < board[0].length && record[xx][yy] == 0 && board[xx][yy] == 'O') {
                record[xx][yy] = 1;
                dfsOnlyWalk(board, record, xx, yy);
            }
        }
    }

    public void dfsOnlyMark(char[][] board, int[][] record, int i, int j) {
        for(int k = 0; k < 4; k++) {
            int xx = i + dx[k];
            int yy = j + dy[k];
            if(xx >= 0 && xx < board.length && yy >= 0 && yy < board[0].length && board[xx][yy] == 'O') {
                board[xx][yy] = 'X';
                dfsOnlyMark(board, record, xx, yy);
            }
        }
    }

    public static void main(String[] args) {
        Solution solution = new Solution();
        /*
        * [["O","X","O","O","X","X"],["O","X","X","X","O","X"],["X","O","O","X","O","O"],["X","O","X","X","X","X"],["O","O","X","O","X","X"],["X","X","O","O","O","O"]]
        * */
        char[][] board = {{'O','X','O','O','X','X'},{'O','X','X','X','O','X'},{'X','O','O','X','O','O'},{'X','O','X','X','X','X'},{'O','O','X','O','X','X'},{'X','X','O','O','O','O'}};
        solution.solve(board);
        for(char[] chars : board) {
            for(char c : chars) {
                System.out.print(c + " ");
            }
            System.out.println();
        }
    }

}

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值