leetcode(NOWCODER)---surrounded-regions

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时间限制：1秒 空间限制：32768K 热度指数：17818
本题知识点： 数组 leetcode
算法知识视频讲解
题目描述

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

/*
    其实就是简单的dfs，关键还是思路：https://www.nowcoder.com/questionTerminal/c159db5028034aa595043a1a220a62dd
    四个边上的O，及与之相连的O都不能变（可以暂时变成#，和其他O区分，最后变回来即可），其他的O肯定是被包围的，变为X
*/

public class Solution {
    public void solve(char[][] board) {
        if(board == null || board.length == 0 || board[0].length == 0){
            return;
        }
        //对四条边上的O进行dfs
        int row = board.length;
        int col = board[0].length;
        for(int i = 0; i < row; i++){
            dfs(board, i, 0);//第0列
            dfs(board, i, col-1);//第col-1列
        }
        for(int i = 0; i < col; i++){
            dfs(board, 0, i);//第0行
            dfs(board, row-1, i);//第row-1行
        }
        //把所有O变为X
        for(int i = 0; i < row; i++){
            for(int j = 0; j < col; j++){
                if(board[i][j] == 'O'){
                    board[i][j] = 'X';
                }
            }
        }
        //把所有#变回O
        for(int i = 0; i < row; i++){
            for(int j = 0; j < col; j++){
                if(board[i][j] == '#'){
                    board[i][j] = 'O';
                }
            }
        }
    }
    private void dfs(char[][] board, int row, int col){
        if(board[row][col] == 'O'){
            board[row][col] = '#';
            if(col > 0){
                dfs(board, row, col-1);
            }
            if(col < board[0].length-1){
                dfs(board, row, col+1);
            }
            if(row > 0){
                dfs(board, row-1, col);
            }
            if(row < board.length-1){
                dfs(board, row+1, col);
            }
        }
    }
}