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
地址：https://oj.leetcode.com/problems/surrounded-regions/
算法：我们必须明确一个事实：所有没有被'X'环绕的'O'必定有一个从边缘的‘O’跟它联通着，所以我的算法就是从四个边缘为'O'的位置开始往内部按BFS的思想将‘O’设置为‘X’。代码：
1 class Solution {
2 public:
3 void solve(vector<vector<char>> &board) {
4 if(board.empty()) return;
5 int m = board.size();
6 vector<vector<bool> > flag;
7 for(int i = 0; i < m; ++i){
8 int mi = board[i].size();
9 flag.push_back(vector<bool>(mi,false));
10 }
11 for(int i = 0; i < board[0].size(); ++i){
12 if(board[0][i] == 'O' && flag[0][i] == false){
13
14 solveCore(board,flag,0,i);
15 }
16 }
17 for(int i = 1; i < m; ++i){
18 int mi = board[i].size();
19 if(mi > 0 && board[i][0] == 'O' && flag[i][0] == false){
20
21 solveCore(board,flag,i,0);
22 }
23 if(mi > 0 && board[i][mi-1] == 'O' && flag[i][mi-1] == false){
24
25 solveCore(board,flag,i,mi-1);
26 }
27 }
28 for(int i = 1; i < board[m-1].size()-1; ++i){
29 if(board[m-1][i] == 'O' && flag[m-1][i] == false){
30
31 solveCore(board,flag,m-1,i);
32 }
33 }
34 for(int i = 0; i < m; ++i){
35 int mi = board[i].size();
36 for(int j = 0; j < mi; ++j){
37 if(board[i][j] == 'O' && !flag[i][j])
38 board[i][j] = 'X';
39 }
40 }
41 }
42 void solveCore(vector<vector<char> > &board, vector<vector<bool> > &flag, int x, int y){
43 queue<pair<int,int> > Q;
44 Q.push(pair<int,int>(x,y));
45 flag[x][y] = true;
46 while(!Q.empty()){
47 pair<int,int> pa = Q.front();
48 int i = pa.first, j = pa.second;
49 if(i > 0 && j < board[i-1].size() && board[i-1][j] == 'O' && flag[i-1][j] == false){
50 flag[i-1][j] = true;
51 Q.push(pair<int,int>(i-1,j));
52 }
53 if(j < board[i].size()-1 && board[i][j+1] == 'O' && flag[i][j+1] == false){
54 flag[i][j+1] = true;
55 Q.push(pair<int,int>(i,j+1));
56 }
57 if(i < board.size()-1 && j < board[i+1].size() && board[i+1][j] == 'O' && flag[i+1][j] == false){
58 flag[i+1][j] = true;
59 Q.push(pair<int,int>(i+1,j));
60 }
61 if(j > 0 && board[i][j-1] == 'O' && flag[i][j-1] == false){
62 flag[i][j-1] = true;
63 Q.push(pair<int,int>(i,j-1));
64 }
65 Q.pop();
66 }
67 }
68 };