leetcode_question_130 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
Method1: O(row*col*min(row,col))
1、把所有边上的不能被X包围的O换成P---O(row*col*min(row,col))，先从走上角开始换，再从右下角开始换，有的时候里面的O其实是和边上的O连通的，但是因为拐弯一次替换不能完成所以就要至少min(row,col)次替换。如果这个弯拐点太大了，这就完蛋了。。。能过Judge Large纯属幸运。。。
2、把里面的被X包围的O换成X---O(row*col)
3、把P换回O---O(row*col)
void solve(vector<vector<char>> &board) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
int row = board.size();
if (row == 0) return;
int col = board[0].size();
if (col == 0) return;
//from left top to right down
for (int j = 0; j < col; ++j)
if (board[0][j] == 'O') board[0][j] = 'P';
for (int i = 0; i < row; ++i)
if (board[i][0] == 'O') board[i][0] = 'P';
for (int i = 1; i < row; ++i)
{
for (int j = 1; j < col; ++j)
{
if ((board[i][j] == 'O') && (board[i][j-1] == 'P' || board[i-1][j] == 'P'))
board[i][j] = 'P';
}
}
//from right down to left top
for (int j = 0; j < col; ++j)
if (board[row-1][j] == 'O') board[row-1][j] = 'P';
for (int i = 0; i < row; ++i)
if (board[i][col-1] == 'O') board[i][col-1] = 'P';
for (int i = row-2; i >= 0; --i)
{
for (int j = col-2; j >= 0; --j)
{
if ((board[i][j] == 'O') && (board[i][j+1] == 'P' || board[i+1][j] == 'P'))
board[i][j] = 'P';
}
}
//ensure
int time = row < col ? row : col;
for (int k = 1; k < time; ++k) {
for (int i = 1; i < row; ++i)
{
for (int j = 1; j < col; ++j)
{
if (board[i][j] == 'O') {
if (board[i][j-1] == 'P' || board[i-1][j] == 'P')
board[i][j] = 'P';
if (j+1 < col && board[i][j+1] =='P')
board[i][j] = 'P';
if (i+1 < row && board[i+1][j] =='P')
board[i][j] = 'P';
}
}
} }
//change O to X
for (int i = 1; i < row; ++i)
{
for (int j = 1; j < col; ++j)
{
if (board[i][j] == 'O')
board[i][j] = 'X';
}
}
//change P to O
for (int i = 0; i < row; ++i)
{
for (int j = 0; j < col; ++j)
{
if (board[i][j] == 'P')
board[i][j] = 'O';
}
}
}
这种方法的缺憾主要在第一步，如果优化的话，就是从矩阵的边界开始找O，只要找到O就从这个O开始BFS搜索把其相邻的O换成P直到相邻的没有O为止。这样就不用这么多次数的O(n^2)了吧。
void changeotop(vector<vector<char>> &board, int i, int j)
{
board[i][j] = 'P';
int row = board.size();
int col = board[0].size();
if(i>0 && board[i-1][j] == 'O')
changeotop(board, i-1, j);
if(j>0 && board[i][j-1] == 'O')
changeotop(board, i, j-1);
if(i+1<row && board[i+1][j] == 'O')
changeotop(board, i+1, j);
if(j+1<col && board[i][j+1] == 'O')
changeotop(board, i, j+1);
}
void solve(vector<vector<char>> &board) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
int row = board.size();
if(row == 0) return;
int col = board[0].size();
if(col == 0) return;
for(int j = 0; j < col; ++j)
if(board[0][j] == 'O')
changeotop(board,0,j);
for(int i = 0; i < row; ++i)
if(board[i][0] == 'O')
changeotop(board,i,0);
for(int j = 0; j < col; ++j)
if(board[row-1][j] == 'O')
changeotop(board,row-1,j);
for(int i = 0; i < row; ++i)
if(board[i][col-1] == 'O')
changeotop(board,i,col);
//change O to X
for(int i = 1; i < row; ++i)
{
for(int j = 1; j < col; ++j)
{
if(board[i][j] == 'O')
board[i][j] = 'X';
}
}
//change P to O
for(int i = 0; i < row; ++i)
{
for(int j = 0; j < col; ++j)
{
if(board[i][j] == 'P')
board[i][j] = 'O';
}
}
}