LeetCode-----把'O'變成'X'

LeetCode—–把’O’變成’X’

注：思路和代碼是參考牛客上Ron大神的，自己沒有寫出來。。。
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
/*
* 所有與四條邊相連的O都保留，其他O都變爲X
* 遍歷四條邊上的O，並深度遍歷與其相連的O，將這些O都轉爲*
* 將剩餘的O變爲X
* 將剩餘的*變爲O
*/

import java.util.*;
public class Solution {
public int rowNum = 0;
public int colNum = 0;
public void solve(char[][] board) {
    if(board == null || board.length <= 0|| board[0].length <= 0){
        return;
    }
    rowNum = board.length;
    colNum = board[0].length;
    for(int i = 0; i < colNum; i++){
        dfs(board, 0, i);
        dfs(board, rowNum-1, i);
    }
    for(int i = 0; i < rowNum; i++){
        dfs(board, i, 0);
        dfs(board, i, colNum-1);
    }
    for(int i = 0; i < rowNum; i++){
        for(int j = 0; j < colNum; j++){
            if(board[i][j] == 'O'){
                board[i][j] = 'X';
            }
        }
    }
    for(int i = 0; i < rowNum; i++){
        for(int j = 0; j < colNum; j++){
            if(board[i][j] == '*'){
                board[i][j] = 'O';
            }
        }
    }
}
private void dfs(char[][] board, int row, int col) {
    // TODO Auto-generated method stub
    if(board[row][col] == 'O'){
        board[row][col] = '*';
        if(row > 1){
            dfs(board, row-1, col);
        }
        if(col > 1){
            dfs(board, row, col-1);
        }
        if(row < rowNum-1){
            dfs(board, row+1, col);
        }
        if(col < colNum-1){
            dfs(board, row, col+1);
        }
    }
}
}