Leetcode: Surrounded regions

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

---

我的最初思路是直接找被包围的O区域，如果这个区域不与边界接触则是被包围区域，但是这种判断要等到dfs完成才能知道是否是被包围区域，所以需要再一次dfs来将这些点变成'X'。所以这样的复杂度太高，Judge Large超时。这个版本的代码包括下面的dfs和changeBoard，以及Solve中被注释掉的部门代码；后来参考了网上的一个代码，将思路转变为先对不被包围的区域做标记'C'，然后再遍历一遍按要求修改。这样就避免了做两遍深度搜索。

代码如下：（有点难看，见谅……）

//
//  SurroundedRegions.cpp
//  SJMcode
//
//  Created by Jiamei Shuai on 13-8-30.
//  Copyright (c) 2013年 Jiamei Shuai. All rights reserved.
//


#include <iostream>
#include <vector>
#include <assert.h>
using namespace std;

class Solution {
public:
    
    void dfs2(vector<vector<char>> &board, int i, int j)
    {
        if(i > board.size()-1 || i < 0 || j > board[0].size()-1 || j < 0)
            return;
        
        if(board[i][j] == 'O')
        {
            board[i][j] = 'C';
            dfs2(board, i+1, j);
            dfs2(board, i-1, j);
            dfs2(board, i, j-1);
            dfs2(board, i, j+1);
        }
            

    }
    
    bool dfs(vector<vector<char>> &board, int i, int j, int height, int width, vector<vector<int>> &isVis, bool &flag) // working but too slow
    {
        //if(board[i][j] == 'X')  return true;
        
        if(i == height-1 || i == 0 || j == width-1 || j == 0)
            flag = false; // 'o' touch the border: means this block cannot be the answer
        
        isVis[i][j] = 1;
        
        
        if(i >= 1 && !isVis[i-1][j] && board[i-1][j] == 'O')
        {
            dfs(board, i-1, j, height, width, isVis, flag); //上
        }
        if(i < (int)board.size()-1 && !isVis[i+1][j] && board[i+1][j] == 'O')
        {
            dfs(board, i+1, j, height, width, isVis, flag); //下
        }
        if(j >= 1 && !isVis[i][j-1] && board[i][j-1] == 'O')
        {
            dfs(board, i, j-1, height, width, isVis, flag); //左
        }
        if(j < (int)board[0].size() && !isVis[i][j+1] && board[i][j+1] == 'O')
        {
            dfs(board, i, j+1, height, width, isVis, flag); //右
        }
        
        return flag;
    }
    
    void changeBoard(vector<vector<char>> &board, int i, int j) // working but too slow
    {
        vector<int> queue;
        board[i][j] = 'X';
        assert(i > 0 && i < board.size()-1 && j > 0 && j < board[0].size()-1);
        if(board[i-1][j] == 'O')
            changeBoard(board, i-1, j);
        if(board[i+1][j] == 'O')
            changeBoard(board, i+1, j);
        if(board[i][j-1] == 'O')
            changeBoard(board, i, j-1);
        if(board[i][j+1] == 'O')
            changeBoard(board, i, j+1);
    }
    
    void solve(vector<vector<char>> &board) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int height = (int)board.size();
        if(height == 0) return;
        int width = (int)board[0].size();
        
//        vector<int> temp(width, 0);
//        
//        vector<vector<int>> isVis(height, temp);
//        bool flag = true;
//        
//        for(int i = 0; i < height; i++)
//        {
//            for(int j = 0; j < width; j++)
//            {
//                if(board[i][j] == 'O' && !isVis[i][j])
//                {
//                    flag = true;
//                    if(dfs(board, i, j, height, width, isVis, flag)) // surround regions
//                    {
//                        changeBoard(board, i, j);  // Find surround region directly may cause runtime error
//                    }
//                }
//            }
//        }
        
        // Change my strategy to mark unsurrounded regions
        
        for(int i = 0; i < height; i++)
        {
            dfs2(board, i, 0);
            dfs2(board, i, width-1);
        }
        
        for(int j = 0; j < width; j++)
        {
            dfs2(board, 0, j);
            dfs2(board, height-1, j);
        }
        
        for(int i = 0; i < height; i++)
        {
            for(int j = 0; j < width; j++)
            {
                if(board[i][j] == 'O') board[i][j] = 'X';
                if(board[i][j] == 'C') board[i][j] = 'O';
            }
        }
        
        // print result
        for(int i = 0; i < height; i++)
        {
            for(int j = 0; j < width; j++)
            {
                cout << board[i][j] << ' ';
            }
            cout << endl;
        }
        
    }
    
};

int main()
{
    vector<vector<char>> board{{'X','X','X','X'},{'X','O','O','X'},{'X','X','O','X'},{'X','O','X','X'}};
    
    
    Solution sln;
    sln.solve(board);
    
    return 0;
}

 此外，提交的代码是不能有输出语句的，否则会报Internal Error。

 BFS也可以做：http://blog.sina.com.cn/s/blog_b9285de20101j1dt.html