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()- || i <  || j > board[].size()- || j < )
             return;

         if(board[i][j] == 'O')
         {
             board[i][j] = 'C';
             dfs2(board, i+, j);
             dfs2(board, i-, j);
             dfs2(board, i, j-);
             dfs2(board, i, j+);
         }

     }

     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- || i ==  || j == width- || j == )
             flag = false; // 'o' touch the border: means this block cannot be the answer

         isVis[i][j] = ;

         if(i >=  && !isVis[i-][j] && board[i-][j] == 'O')
         {
             dfs(board, i-, j, height, width, isVis, flag); //上
         }
         if(i < (int)board.size()- && !isVis[i+][j] && board[i+][j] == 'O')
         {
             dfs(board, i+, j, height, width, isVis, flag); //下
         }
         if(j >=  && !isVis[i][j-] && board[i][j-] == 'O')
         {
             dfs(board, i, j-, height, width, isVis, flag); //左
         }
         if(j < (int)board[].size() && !isVis[i][j+] && board[i][j+] == 'O')
         {
             dfs(board, i, j+, 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 >  && i < board.size()- && j >  && j < board[].size()-);
         if(board[i-][j] == 'O')
             changeBoard(board, i-, j);
         if(board[i+][j] == 'O')
             changeBoard(board, i+, j);
         if(board[i][j-] == 'O')
             changeBoard(board, i, j-);
         if(board[i][j+] == 'O')
             changeBoard(board, i, j+);
     }

     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 == ) return;
         int width = (int)board[].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 = ; i < height; i++)
         {
             dfs2(board, i, );
             dfs2(board, i, width-);
         }

         for(int j = ; j < width; j++)
         {
             dfs2(board, , j);
             dfs2(board, height-, j);
         }

         for(int i = ; i < height; i++)
         {
             for(int j = ; 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 = ; i < height; i++)
         {
             for(int j = ; 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 ;
 }

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