November 26, 2019

130. Surrounded Regions

这道题和这两道题是一样的思路： 200. Number of Islands
286. Walls and Gates

Loop the out layer, if found any ‘O’, do DFS/BFS, mark any ‘O’ with ‘B’
Loop the matrix again, flip ‘O’ to ‘X’, and flip ‘B’ to ‘O’.

Sol 1. DFS

Time = O(mn), space = O(mn) (递归栈的大小)

class Solution {
    private int[][] dirs = {`{-1, 0}, {1, 0}, {0, -1}, {0, 1}`};

    public void solve(char[][] board) {
        if (board == null || board.length == 0 || board[0].length == 0) {
            return;
        }
        for (int i = 0; i < board.length; i++) {
            if (board[i][0] == 'O') {
                dfsHelper(i, 0, board);
            }
            if (board[i][board[0].length - 1] == 'O') {
                dfsHelper(i, board[0].length - 1, board);
            }
        }
        for (int j = 0; j < board[0].length; j++) {
            if (board[0][j] == 'O') {
                dfsHelper(0, j, board);
            }
            if (board[board.length - 1][j] == 'O') {
                dfsHelper(board.length - 1, j, board);
            }
        }
        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';
                }
            }
        }
    }
    
    private void dfsHelper(int i, int j, char[][] board) {
        if (i < 0 || i > board.length - 1 || j < 0 || j > board[0].length - 1 || board[i][j] != 'O') { // 注意这里剪枝要剪掉X和B
            return;
        }
        board[i][j] = 'B';
        for (int[] dir : dirs) {
            dfsHelper(i + dir[0], j + dir[1], board);
        }
    }
}

Sol 2. BFS

Time = O(mn), space = O(mn)(?) (Queue的大小)

class Solution {
    
    public void solve(char[][] board) {
        if (board == null || board.length == 0 || board[0].length == 0) {
            return;
        }
        
        int[][] dirs = {`{-1, 0}, {1, 0}, {0, -1}, {0, 1}`};
        Deque<int[]> queue = new ArrayDeque<>();
        for (int i = 0; i < board.length; i++) {
            if (board[i][0] == 'O') {
                queue.offerLast(new int[]{i, 0});
            }
            if (board[i][board[0].length - 1] == 'O') {
                queue.offerLast(new int[]{i, board[0].length - 1});
            }
        }
        for (int j = 0; j < board[0].length; j++) {
            if (board[0][j] == 'O') {
                queue.offerLast(new int[]{0, j});
            }
            if (board[board.length - 1][j] == 'O') {
                queue.offerLast(new int[]{board.length - 1, j});
            }
        }
        
        while (!queue.isEmpty()) {
            int[] cur = queue.pollFirst();
            board[cur[0]][cur[1]] = 'B';
            for (int[] dir : dirs) {
                int x = cur[0] + dir[0];
                int y = cur[1] + dir[1];
                if (x < 0 || x > board.length - 1 || y < 0 || y > board[0].length - 1 || board[x][y] != 'O') {
                    continue;
                }
                queue.offerLast(new int[]{x, y});
            }
        }        
        
        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';
                }
            }
        }
    }
}

Sol 3. Union Find

TODO