leetcode 130.Surrounded Regions

leetcode 130.Surrounded Regions

题目：

Given a 2D board containing \'X\' and \'O\' (the letter O), capture all regions surrounded by \'X\'.

A region is captured by flipping all \'O\'s into \'X\'s in that surrounded region.

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

Explanation:

Surrounded regions shouldn’t be on the border, which means that any \'O\' on the border of the board are not flipped to \'X\'. Any \'O\' that is not on the border and it is not connected to an \'O\' on the border will be flipped to \'X\'. Two cells are connected if they are adjacent cells connected horizontally or vertically.

解法：

拿到这个题首先要理清题意，这个题是让我们将一个矩形图中除边界上的\'O\'以及与边界上的\'O\'直接相连的\'O\'外的所有\'O\'都转化为\'X\'

对于矩形图的问题，我们一般考虑使用 BFS 算法（也即宽度优先搜索），从图的左上角往右下角进行搜索，但是这个题我们不需要做太多的无用功，因为example 1中已经明显给出了我们要找的满足条件的情况，所以我们只需要从边界出发，搜边界上的点即可。

同时我们需要确保：

• 在搜索过程中不走重复的道路，也即我们需要将已经搜索过的\'O\'替换为别的字符(这里我使用的\'$\')，这样可以减少我们搜索的时间复杂度;
• 当我们搜索所有完成时候，将没有被替换的\'O\'替换为\'X\'(这是因为没有被搜索过的\'O\'都是边界上的\'O\'所不可达的)，将\'$\'置换为\'O\';
• 某一个\'O\'所在位置的四个方向都要进行搜索，需要保证我们的搜索不能越界

代码：

class Solution {
public:
    void solve(vector<vector<char> >& board) {
        for (int i = 0; i < board.size(); ++i) {
            for (int j = 0; j < board[i].size(); ++j) {
                if ((i == 0 || i == board.size() - 1 || j == 0 || j == board[i].size() - 1) && board[i][j] == \'O\')
                    solveDFS(board, i, j);
            }
        }
        for (int i = 0; i < board.size(); ++i) {
            for (int j = 0; j < board[i].size(); ++j) {
                if (board[i][j] == \'O\') board[i][j] = \'X\';
                if (board[i][j] == \'$\') board[i][j] = \'O\';
            }
        }
    }
    void solveDFS(vector<vector<char> > &board, int i, int j) {
        if (board[i][j] == \'O\') {
            board[i][j] = \'$\';
            if (i > 0 && board[i - 1][j] == \'O\') 
                solveDFS(board, i - 1, j);
            if (j < board[i].size() - 1 && board[i][j + 1] == \'O\') 
                solveDFS(board, i, j + 1);
            if (i < board.size() - 1 && board[i + 1][j] == \'O\') 
                solveDFS(board, i + 1, j);
            if (j > 1 && board[i][j - 1] == \'O\') 
                solveDFS(board, i, j - 1);
        }
    }
};