LeetCode 130. Surrounded Regions

原题链接在这里：https://leetcode.com/problems/surrounded-regions/

题目：

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

Explanation:

Surrounded regions shouldn’t be on the border, which means that any 'O' on the border of the board are not flipped to 'X'. Any 'O' that is not on the border and it is not connected to an 'O' on the border will be flipped to 'X'. Two cells are connected if they are adjacent cells connected horizontally or vertically.

题解：

第一行和最后一行，第一列和最后一列，若是有'O'才能保留，并且和它们相连的'O'才能同时保留。

所以从这四条边上的'O' 点开始做BFS 或者 DFS, 把所有相连的点都变成 '#'.

最后扫描一遍board, 先判定这点若还是 'O', 说明这个'O'没有与边界相连，即是被包围的'O', 改为'X'.

后判定这点若是'#', 说明这个'O'与边界相连，应该保存，改回'O'.

Note: 无论是BFS还是DFS都应该有个used来check 之前是否visit 过这个cell. 否则就会无限循环. 这里可以先在board上更改, 再进queue, 如果出queue再更改就会有重复进queue的情况.

Time Complexity: O(m*n).

Space: O(m*n).

AC Java:

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;
        }
        
        int m = board.length;
        int n = board[0].length;
        for(int j = 0; j<n; j++){
            if(board[0][j] == 'O'){
                // bfs(board, 0, j);
                dfs(board, 0, j);
            }
            
            if(board[m-1][j] == 'O'){
                // bfs(board, m-1, j);
                dfs(board, m-1, j);
            }
        }
        
        for(int i = 0; i<m; i++){
            if(board[i][0] == 'O'){
                // bfs(board, i, 0);
                dfs(board, i, 0);
            }
            
            if(board[i][n-1] == 'O'){
                // bfs(board, i, n-1);
                dfs(board, i, n-1);
            }
        }
        
        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 i ,int j){
        if(i<0 || i>=board.length || j<0 || j>=board[0].length || board[i][j]!='O'){
            return;
        }
        
        board[i][j] = '#';
        for(int [] dir : dirs){
            dfs(board, i+dir[0], j+dir[1]);
        }
    }
    
    private void bfs(char [][] board, int i, int j){
        int m = board.length;
        int n = board[0].length;
        
        if(i<0 || i>=m || j<0 || j>=n){
            return;
        }
    
        LinkedList<Integer> que = new LinkedList<Integer>();
        board[i][j] = '#';
        que.add(getIndex(i, j, n));
        
        while(!que.isEmpty()){
            int cur = que.poll();
            int p = cur/n;
            int q = cur%n;
            
            for(int [] dir : dirs){
                int xp = p + dir[0];
                int xq = q + dir[1];
                if(xp>=0 && xp<m && xq>=0 && xq<n && board[xp][xq] == 'O'){
                    board[xp][xq] = '#';
                    que.add(getIndex(xp, xq, n));
                }
            }
        }
    }
    
    private int getIndex(int i, int j, int n){
        return i*n+j;
    }
}

类似Number of Enclaves, Number of Islands.