[LeetCode] 130. Surrounded Regions

Given a 2D board containing 'X' and 'O' (the letter O), capture all regions surrounded by 'X'.

A region is captured by flipping all 'O's into 'X's in that surrounded region.

Example:

X X X X
X O O X
X X O X
X O X X

After running your function, the board should be:

X X X X
X X X X
X X X X
X O X X

被围绕的区域。题意是给一个二维矩阵，用大写的X和O（欧，不是零）填满。请找到所有被 'X' 围绕的区域，并将这些区域里所有的 'O' 用 'X' 填充。题意很简单但是注意一个边界条件。题目要你将所有被包围的O用X填充，但是不包括矩阵边界的O。意思是如果矩阵边界上有O，或者矩阵内部的O实际是跟边界上的O有竖直或者水平能连接起来的（类似200题的岛的概念），你就不能将这种O置换成X。

思路还是DFS，但是首先要扫描矩阵的四条边，看看四个border上是否有O，如果有，则需要DFS遍历，找出所有跟这个边界上的O扯上关系的O并且把他们标记成一个别的东西。我这里是标记成1。标记说明这些坐标是不能被改成X的。接下来再次遍历整个矩阵，如果当前坐标上是O，则将这个坐标变成X；如果遇到1，则将其变成O。

时间O(mn)

空间O(n)

Java实现

class Solution {
    public void solve(char[][] board) {
        // corner case
        if (board == null || board.length == 0 || board[0].length == 0) {
            return;
        }

        // normal case
        int m = board.length - 1;
        int n = board[0].length - 1;
        // first row and last row
        for (int i = 0; i <= m; i++) {
            if (board[i][0] == 'O') {
                helper(board, i, 0);
            }
            if (board[i][n] == 'O') {
                helper(board, i, n);
            }
        }
        // first column and last column
        for (int i = 0; i <= n; i++) {
            if (board[0][i] == 'O') {
                helper(board, 0, i);
            }
            if (board[m][i] == 'O') {
                helper(board, m, i);
            }
        }

        // the rest
        for (int i = 0; i <= m; i++) {
            for (int j = 0; j <= n; j++) {
                if (board[i][j] == 'O') {
                    board[i][j] = 'X';
                } else if (board[i][j] == '1') {
                    board[i][j] = 'O';
                }
            }
        }
    }

    private void helper(char[][] board, int r, int c) {
        if (r < 0 || c < 0 || r > board.length - 1 || c > board[0].length - 1 || board[r][c] != 'O') {
            return;
        }
        board[r][c] = '1';
        helper(board, r + 1, c);
        helper(board, r - 1, c);
        helper(board, r, c - 1);
        helper(board, r, c + 1);
    }
}

flood fill题型总结

LeetCode 题目总结