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
广度优先搜索
我们只需要从最外围的一圈中找到O,并且对其用广度优先搜索找到与外围O连通的路径,标记为'E',所有非E的都为X
class Solution {
public:
int m ,n;
void solve(vector> &board) {
if (board.size() == 0) return;
m = board.size();
n = board[0].size();
for (int i = 0; i < n; i++) {
bfs(board, 0 , i);
bfs(board, m-1, i);
}
for (int j = 1; j < m-1; j++) {
bfs(board, j, 0);
bfs(board, j, n-1);
}
for (int i = 0; i < m; i++) {
for (int j = 0; j > &board, int i, int j) {
queue q;
visit(board,i,j,q);
while(!q.empty()) {
int cur = q.front();
q.pop();
const int x = cur/n;
const int y = cur%n;
visit(board, x-1, y, q);
visit(board, x, y-1, q);
visit(board, x+1, y, q);
visit(board, x, y+1, q);
}
}
void visit(vector> &board, int i, int j, queue &q) {
if(i < 0 || i > m-1 || j <0 || j > n-1 || board[i][j] != 'O')
return;
board[i][j] = 'E';
q.push(i*n+j);
}
};
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