LeetCode:Surrounded Regions

题目如下：（题目链接）

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

分析：先找到边缘的O，与边缘的O连通的所有O均不用转化成X（这里连通的意思是从边缘的O开始有一条全是O的路径可以到达该点），其他的所有O都变成X。我们只要从边缘的O开始dfs标记所有与它连通的O，然后再把所有未标记的O变成X。

下面的代码中是使用了一个标记数组。如果想节省空间，可以先把与边缘O连通的O先变成#，然后再把剩余的O变为X，再把#变成O。本文地址

 1  
 2 class Solution {
 3 public:
 4     struct Qnode
 5     {
 6         int i_;
 7         int j_;
 8         Qnode(int x,int y):i_(x),j_(y){};
 9         Qnode(){};
10     };
11     void solve(vector<vector<char>> &board) {
12         // IMPORTANT: Please reset any member data you declared, as
13         // the same Solution instance will be reused for each test case.
14         int row = board.size();
15         if(row == 0)return;
16         int colum = board[0].size();
17         queue<Qnode> myqueue;
18         bool visited[row][colum];
19         memset(visited, 0, sizeof(visited));
20         //首先标记所有没有被包围的O
21         for(int i = 0; i < row; i++)
22         {
23             for(int j = 0; j < colum; j++)
24             {//首先找到四周的O，和这个O连成一片的O都不用转换
25                 if(i == 0 || i == row - 1 || j == 0 || j == colum - 1)
26                     if(visited[i][j] == false && board[i][j] == 'O')
27                         {
28                             visited[i][j] = true;
29                             myqueue.push(Qnode(i,j));
30                             while(myqueue.empty() == false)
31                             {
32                                 Qnode tmp = myqueue.front();
33                                 myqueue.pop();
34                                 //board[tmp.i_][tmp.j_] = 'X';
35                                 if(tmp.i_-1>=0 && board[tmp.i_-1][tmp.j_] == 'O'
36                                     &&visited[tmp.i_-1][tmp.j_] == false)
37                                 {//上
38                                     visited[tmp.i_-1][tmp.j_] = true;
39                                     myqueue.push(Qnode(tmp.i_-1, tmp.j_));
40                                 }
41                                 if(tmp.i_+1 < row && board[tmp.i_+1][tmp.j_] == 'O'
42                                     &&visited[tmp.i_+1][tmp.j_] == false)
43                                 {//下
44                                     visited[tmp.i_+1][tmp.j_] = true;
45                                     myqueue.push(Qnode(tmp.i_+1, tmp.j_));
46                                 }
47                                 if(tmp.j_-1 >= 0 && board[tmp.i_][tmp.j_-1] == 'O'
48                                     &&visited[tmp.i_][tmp.j_-1] == false)
49                                 {//左
50                                     visited[tmp.i_][tmp.j_-1] = true;
51                                     myqueue.push(Qnode(tmp.i_, tmp.j_-1));
52                                 }
53                                 if(tmp.j_+1 < colum && board[tmp.i_][tmp.j_+1] == 'O'
54                                     &&visited[tmp.i_][tmp.j_+1] == false)
55                                 {//右
56                                     visited[tmp.i_][tmp.j_+1] = true;
57                                     myqueue.push(Qnode(tmp.i_, tmp.j_+1));
58                                 }
59                             }
60                         }
61             }
62         }
63         //把未被标记的O转换成X
64         for(int i = 0; i < row; i++)
65             for(int j = 0; j < colum; j++)
66                 if(board[i][j] == 'O' && visited[i][j] == false)
67                     board[i][j] = 'X';
68         
69     }
70 };