Leetcode 130 Surrounded Regions DFS

将内部的O点变成X

input

X X X X
X O O X
X X O X
X O X X

output

X X X X
X X X X
X X X X
X O X X

DFS的基本框架是

void dfs(int now,int d){
      if(终止条件) {
          做相应的操作;
           return;
      }
       for(遍历所有now点的相邻点next){
            if(!visit[next]) {
                 访问每个没有访问过的点;
                 做相应的操作;
                 dfs(next, d + 1);
            }
         }         
  }
    

DFS图所有边上的点，将边上的O以及其联通域变为T，然后将内部的O变为X,外部的T变为O，本质是种子填充算法。

此题对于is_in这个函数要注意， 不要用(x < m) && (x >= 0) && (y < n) && (y >= 0) 会堆栈溢出！！原因是会出现边上的点全是O。。。

class Solution {
public:
    
    int m, n;
    bool is_in(int x, int y)
    {
        return (x < m-1) && (x >= 1) && (y < n-1) && (y >= 1);
    }

    void dfs(std::vector<std::vector<char>> &board,int x,int y)
    {
        //if (!(is_in(x, y) && board[x][y] == 'O')) return;
        //printf("%d %d
", x, y);
        board[x][y] = 'T';
        int dir[4][2] = { { 1, 0 }, { -1, 0 }, { 0, 1 }, { 0, -1 } };
        for (int i = 0; i < 4; ++i){
            int tx = x + dir[i][0];
            int ty = y + dir[i][1];
            if (is_in(tx, ty) && board[tx][ty] == 'O')
            {
                dfs(board, tx, ty);
            }
        }
    }

    void change(std::vector<std::vector<char>> &board)
    {
        for (int i = 0; i < m;++i){
            for (int j = 0; j < n;++j){
                if (board[i][j] == 'T') board[i][j] = 'O';
                else if (board[i][j] == 'O') board[i][j] = 'X';
                else;
            }
        }
    }

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