leetcode21 surrounded regions

题目描述

现在有一个仅包含‘X’和‘O’的二维板，请捕获所有的被‘X’包围的区域

捕获一个被包围区域的方法是将被包围区域中的所有‘O’变成‘X’

例如

X X X X↵X O O X↵X X O X↵X O X X

执行完你给出的函数以后，这个二维板应该变成：

X X X X↵X X X X↵X X X X↵X O X X

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
class Solution {
public:
    void solve(vector<vector<char>> &board) {
        if(board.empty())
            return;
        int rows = board.size();
        int cols = board[0].size();
         
        if(rows==0 || cols==0)
            return;
         
        for(int j=0;j<cols;j++)
        {
            DFS(board, 0, j);
            DFS(board, rows-1, j);         }                  for(int i=0;i<rows;i++)         {             DFS(board, i, 0);             DFS(board, i, cols-1);         }                  for(int i=0;i<rows;i++)             for(int j=0;j<cols;j++)                 if(board[i][j] == 'O')                     board[i][j] = 'X';                  for(int i=0;i<rows;i++)             for(int j=0;j<cols;j++)                 if(board[i][j] == '*')                     board[i][j] = 'O';
    }
    void DFS(vector<vector<char> > &board, int r, int c)
    {
        if(board[r][c] == 'O')
        {
            board[r][c] = '*';             int rows = board.size();             int cols = board[0].size();                          if(r > 1)                 DFS(board, r-1, c);             if(r < rows-2)                 DFS(board, r+1, c);             if(c > 1)                 DFS(board, r, c-1);             if(c < cols-2)                 DFS(board, r, c+1);                     }     }
};