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

1、把所有邊上的不能被X包圍的O換成P---O(row*col*min(row,col))，先從走上角開始換，再從右下角開始換，有的時候裡面的O其實是和邊上的O連通的，但是因為拐彎一次替換不能完成所以就要至少min(row,col)次替換。如果這個彎拐點太大了，這就完蛋了。。。能過Judge Large純屬幸運。。。

2、把裡面的被X包圍的O換成X---O(row*col)

3、把P換回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';
        }
    }

}

這種方法的缺憾主要在第一步，如果優化的話，就是從矩陣的邊界開始找O，只要找到O就從這個O開始BFS搜尋把其相鄰的O換成P直到相鄰的沒有O為止。這樣就不用這麼多次數的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';
			}
		}
    }