leetcode 130 Surrounded Regions(BFS)

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

题解：这道题测试数据卡的很严。一开始用dfs爆栈了，后来改成bfs又超时了。。最后改进了下：先查边上有没有‘O’，和边上连着的肯定不会被围住。把这些都标记出来以后，剩下的里面的‘O’必然都要变成‘X’。

class Solution {
public:
int n,m;
int dx[4]={0,0,1,-1};
int dy[4]={1,-1,0,0};
vector<vector<char> >flag;

void bfs(vector<vector<char> >& board,int x,int y){
queue<pair<int,int> >q;
q.push(make_pair(x,y));
flag[x][y]=1;

while(!q.empty()){
int qx=q.front().first;
int qy=q.front().second;
q.pop();
for(int i=0;i<4;i++){
int tx=qx+dx[i];
int ty=qy+dy[i];
if(tx>=0&&tx<n&&ty>=0&&ty<m&&flag[tx][ty]==0&&board[tx][ty]=='O'){
q.push(make_pair(tx,ty));
flag[tx][ty]=1;
}
}
}
}

void solve(vector<vector<char> >& board) {
if(board.empty()){
return ;
}
n=board.size();
m=board[0].size();
flag.resize(n);
for(int i=0;i<n;i++){
flag[i].resize(m,0);
}

for(int i=0;i<m;i++){
if(board[0][i]=='O'&&flag[0][i]==0){
bfs(board,0,i);
}
if(board[n-1][i]=='O'&&flag[n-1][i]==0){
bfs(board,n-1,i);
}
}

for(int i=1;i<=n-2;i++){
if(board[i][0]=='O'&&flag[i][0]==0){
bfs(board,i,0);
}
if(board[i][m-1]=='O'&&flag[i][m-1]==0){
bfs(board,i,m-1);
}
}

for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
if(board[i][j]=='O'&&flag[i][j]==0){
board[i][j]='X';
}
}
}
}
};