Leetcode130. 被围绕的区域

题目传送：​​https://leetcode.cn/problems/surrounded-regions/​​

运行效率：

代码如下：

// 用于记录需要保留的'O',  如果boardFlag[i][j]=true;那就说明坐标[i][j]位置的'O'需要保留，而不需要替换成‘X’
  boolean[][] boardFlag = null;

    public void solve(char[][] board) {
        if (boardFlag == null) {
            boardFlag = new boolean[board.length][board[0].length];
        }
        //1、先从最上面一行开始，计算需要保留的’O‘
        for (int i = 0; i < board[0].length; i++) {
            dfs(board, 0, i);
        }
        //2、从最下面一行开始，计算需要保留的’O‘
        for (int i = 0; i < board[0].length; i++) {
            dfs(board, board.length - 1, i);
        }
        //3、从最左边一列开始，计算需要保留的’O‘
        for (int i = 0; i < board.length; i++) {
            dfs(board, i, 0);
        }
        //4、从最右边一列开始，计算需要保留的’O‘
        for (int i = 0; i < board.length; i++) {
            dfs(board, i, board[0].length-1);
        }
  //上面的4个for循环已经把需要保留的‘O’的位置标记在boardFlag 中了
        //从左到右，  从上到下的遍历
        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[0].length; j++) {                 //把不需要保留的‘O’替换成'X'
                if (board[i][j] == 'O' && !boardFlag[i][j]) {
                    board[i][j] = 'X';
                }
            }
        }
    }

    public void dfs(char[][] board, int i, int j) {
        //处理边界情况
        if (board[i][j] != 'O') {
            return;
        }
        boardFlag[i][j] = true;
        //上
        if (i - 1 >= 0 && !boardFlag[i - 1][j]) {
            dfs(board, i - 1, j);
        }
        //下
        if (i + 1 <= board.length - 1 && !boardFlag[i + 1][j]) {
            dfs(board, i + 1, j);
        }
        //左
        if (j - 1 >= 0 && !boardFlag[i][j - 1]) {
            dfs(board, i, j - 1);
        }
        //右
        if (j + 1 <= board[0].length - 1 && !boardFlag[i][j + 1]) {
            dfs(board, i, j + 1);
        }
    }