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给定一个由字符‘X‘和‘O’组成的矩阵，消除所有被‘X’包围的‘O’

X X X X
X O O X
X X O X
X O X X
变成
X X X X
X X X X
X X X X
X O X X
思路：有点像围棋里一块区域是否有气，只有有气的才能存活下来。即只有‘O’到达了四条边界才可以存活，因此从四条边界开始做广搜，将所有搜索到的‘O‘赋值为‘1’，然后遍历整个矩阵将所有之前没有被赋值的‘O’赋值为‘X’，将‘1’还原为‘O’即可。

class Solution {
public:
    void solve(vector<vector<char>> &board) {
        if (board.empty())
            return;
        int width = board[0].size();
        int height = board.size();
        for (size_t i = 0; i < width; i++)
        {
            if (board[0][i] == ‘O‘)
                bfs(0, i, board);
            if (board[height - 1][i] == ‘O‘)
                bfs(height-1, i, board);
        }
        for (size_t i = 0; i < height; i++)
        {
            if (board[i][0] == ‘O‘)
                bfs(i, 0, board);
            if (board[i][width - 1] == ‘O‘)
                bfs(i, width - 1, board);
        }
        for (size_t i = 0; i < height; i++)
        {
            for (size_t j = 0; j < width; j++)
            {
                if (board[i][j] == ‘1‘)
                    board[i][j] = ‘O‘;
                else if (board[i][j] == ‘O‘)
                {
                    board[i][j] = ‘X‘;
                }
            }
        }
    }
    void bfs(int x,int y, vector<vector<char> >&board)
    {
        board[x][y] = ‘1‘;
        queue<Point> qPoint;
        qPoint.push(Point(x,y));
        Point tmpPoint=Point(0,0);
        int xBound = board.size();
        int yBound = board[0].size();
        while (!qPoint.empty())
        {
            tmpPoint = qPoint.front();
            x = tmpPoint.x;
            y = tmpPoint.y;
            if (isInside(x+1, y, xBound, yBound) && board[x+1][y] == ‘O‘)
            {
                qPoint.push(Point(x+1, y));
                board[x+1][y] = ‘1‘;
            }
            if (isInside(x-1, y, xBound, yBound) && board[x-1][y] == ‘O‘)
            {
                qPoint.push(Point(x-1, y));
                board[x-1][y] = ‘1‘;
            }
            if (isInside(x, y+1, xBound, yBound) && board[x][y+1] == ‘O‘)
            {
                qPoint.push(Point(x, y+1));
                board[x][y+1] = ‘1‘;
            }
            if (isInside(x, y - 1, xBound, yBound) && board[x][y - 1] == ‘O‘)
            {
                qPoint.push(Point(x, y - 1));
                board[x][y - 1] = ‘1‘;
            }
            qPoint.pop();
        }
    }
    bool isInside(int x, int y, int xBound, int yBound)
    {
        return x >= 0 && x < xBound && y >= 0 && y < yBound;
    }
    struct Point
    {
        int x;
        int y;
        Point(int a, int b) :x(a), y(b){}
    };
};

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原文地址：http://www.cnblogs.com/flyjameschen/p/fcdadc64b3ea86ad9861337babbd17ec.html