130. Surrounded Regions

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.

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

题意：
将所有被X包住的O都变成X，但是和边缘相邻的O不算被包围

思路：

DFS：只从上下左右四个边缘开始递归，将遇到的‘O’全部修改为一个特殊的字符例如‘$’，最后遍历一遍数组，将‘$’改为‘O’，非‘$’的位置改为‘X’。

注：

1. 只有元素为‘O’时才开始递归
2. 对于递归终止的条件，最好在进入递归前用if限制，而不是在进入后用return终止。

下面两种实现方式相差80ms左右，二他们唯一的区别只是定义递归是用的dir的位置不一样，92ms的在dfs函数中定义。12ms的定义为全局变量。因为如果定义在dfs函数中，每次递归调用dfs都要初始化一次dir。这种在递归函数中使用的变量，尽量定义为全局的。不要每次调用时都要重新声明初始化。

解法1：92ms

class Solution {
public:
    void solve(vector<vector<char>>& board) {
        if (board.empty()) return;
        int m = board.size(), n = board[0].size();
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (i == 0 || i == m - 1 || j == 0 || j == n - 1) {
                    if (board[i][j] == 'O') dfs(board, i , j);
                }
            }   
        }
        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';
            }
        }
    }

    void dfs(vector<vector<char>> &board, int x, int y) {
        int m = board.size(), n = board[0].size();
        vector<vector<int>> dir{{0,-1},{-1,0},{0,1},{1,0}}; //导致耗时的原因
        board[x][y] = '$';
        for (int i = 0; i < dir.size(); ++i) {
            int dx = x + dir[i][0], dy = y + dir[i][1];
            if (dx >= 0 && dx < m && dy > 0 && dy < n && board[dx][dy] == 'O') {
                dfs(board, dx, dy);
            }
        }
    }
};

解法2 ：12ms

int dx[] = {1,-1,0,0};
int dy[] = {0,0,1,-1};
class Solution {

public:
    void solve(vector<vector<char>>& board) {
        if(!board.size()) return;
        int n = board.size();
        int m = board[0].size();

        for(int i = 0;i<n;++i){
            for(int j = 0;j < m;++j){
                if (i == 0 || i == n - 1 || j == 0 || j == m - 1) {
                    if(board[i][j] == 'O')
                        dfs(i,j,board);
                }
            }
        }

        for(int i = 0;i<n;++i){
            for(int j = 0;j < m;++j){
                if(board[i][j]=='O') board[i][j]='X';
                if(board[i][j]=='$') board[i][j]='O';
            }
        }
        return;
    }

    void dfs(int i,int j,vector<vector<char> >& board){
        int n = board.size(), m = board[0].size();
        board[i][j]='$';
        for(int k = 0;k < 4;++k){
            int nx = dx[k] + i;
            int ny = dy[k] + j;
            if(nx >= 0 && nx < n &&ny > 0&& ny < m && board[nx][ny] == 'O') //注意必须ny>0,ny=0会报错，这是这道题的bug
                dfs(nx,ny,board);
        }
        return;
    }
};