Leetcode 130. Surrounded Regions

二维棋盘O变X策略

最新推荐文章于 2020-09-27 18:35:00 发布

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本文介绍了一种解决二维棋盘中被X包围的O转换为X的问题的方法。通过两次遍历矩阵实现：首次遍历从边界上的O开始进行深度优先搜索并标记；二次遍历则将未被标记的O转换为X，而标记过的O还原。

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.

For 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

题目的意思是：把 X 围着的 O 变为 X。第一次遍历矩阵的时候，我们可以从四个边开始往里找，把 O 变为 #。

第二次遍历的时候可以把 # 变为 O，因为它们与四个边相连，再把 O 变为 X。

public void solve(char[][] board) {
        if (board == null || board.length == 0 || board[0].length == 0) return;
        int m = board.length, n = board[0].length;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (i == 0 || i == m - 1 || j == 0 || j == n - 1) {
                    if (board[i][j] == 'O') dfs(board, i, j);
                }
            }
        }
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; ++j) {
                if (board[i][j] == 'O') board[i][j] = 'X';
                if (board[i][j] == '#') board[i][j] = 'O';
            }
        }
    }
    
    private void dfs(char[][] board, int x, int y) {
        int m = board.length, n = board[0].length;
        int[][] dirs = new int[][] {{0, -1}, {0, 1}, {1, 0}, {-1, 0}};
        board[x][y] = '#';
        for (int i = 0; i < dirs.length; i++) {
            int dx = x + dirs[i][0], dy = y + dirs[i][1];
            if (dx > 0 && dx < m && dy > 0 && dy < n && board[dx][dy] == 'O') dfs(board, dx, dy);
        }
    }