leetcode 130. Surrounded Regions

最新推荐文章于 2023-12-01 14:27:39 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’区域捕获问题的算法。通过使用广度优先搜索（BFS），从棋盘的边界开始遍历，将符合条件的‘O’标记为临时状态‘B’，最后将所有未被标记的‘O’替换为‘X’，实现捕获效果。

Question

Given a 2D board containing ‘X’ and ‘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

code

public void solveC(char[][] board) {
        if (board == null || board.length == 0 || board[0].length == 0) {
            return;
        }
        for (int i = 0; i < board[0].length; i++) {
            bfsC(board, 0, i);
            bfsC(board, board.length - 1, i);
        }
        for (int j = 0; j < board.length; j++) {
            bfsC(board, j, 0);
            bfsC(board, j, board[0].length - 1);
        }
        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[0].length; j++) {
                if (board[i][j] == 'O') {
                    board[i][j] = 'X';
                } else if (board[i][j] == 'B') {
                    board[i][j] = 'O';
                }
            }
        }

    }

    public void bfsC(char[][] board, int i, int j) {
        int rows = board.length;
        int cols = board[0].length;

        Queue<Integer> queues = new LinkedList<>();
        queues.add(i * cols + j);
        while (!queues.isEmpty()) {
            Integer t = queues.poll();
            int irow = t / cols;
            int icol = t % cols;
            if (board[irow][icol] != 'O') {
                continue;
            }
            board[irow][icol] = 'B';

            if (irow > 0) {
                queues.add(t - cols);
            }
            if (irow < rows - 1) {
                queues.add(t + cols);
            }
            if (icol > 0) {
                queues.add(t - 1);
            }
            if (irow < cols - 1) {
                queues.add(t + 1);
            }
        }
    }