[leetcode]130. Surrounded Regions

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本文介绍了一种解决LeetCode上围绕区域问题的方法。该问题要求在二维棋盘中捕获所有被 'X' 包围的 'O' 区域，并将其翻转为 'X'。解决方案首先检查矩阵边界上的 'O' 并将其及其相邻 'O' 修改为临时标记，再将剩余的 'O' 转换为 'X'，最后恢复边界标记。

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题目链接：https://leetcode.com/problems/surrounded-regions/#/description

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.

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

First, check the four border of the matrix. If there is a element is
'O', alter it and all its neighbor 'O' elements to '1'.

Then ,alter all the 'O' to 'X'
At last,alter all the '1' to 'O'

For example:

         X X X X           X X X X             X X X X
         X X O X  ->       X X O X    ->       X X X X
         X O X X           X 1 X X             X O X X
         X O X X           X 1 X X             X O X X

class Solution {
public:
    void solve(vector<vector<char>>& board) {
        int i,j;
        int row=board.size();
        if(!row)
            return;
        int col=board[0].size();

        for(i=0;i<row;i++)
        {
            check(board,i,0,row,col);
            if(col>1)
                check(board,i,col-1,row,col);
        }
        for(j=1;j+1<col;j++)
        {
            check(board,0,j,row,col);
            if(row>1)
                check(board,row-1,j,row,col);
        }
        for(i=0;i<row;i++)
        {
            for(j=0;j<col;j++)
            {
                if(board[i][j]=='O')
                    board[i][j]='X';
            }
        }
        for(i=0;i<row;i++)
        {
            for(j=0;j<col;j++)
            {
                if(board[i][j]=='1')
                    board[i][j]='O';
            }
        }
    }
    void check(vector<vector<char>>& vec,int i,int j,int row,int col)
    {
        if(vec[i][j]=='O')
        {
            vec[i][j]='1';
            if(i>1)
                check(vec,i-1,j,row,col);
            if(j>1)
                check(vec,i,j-1,row,col);
            if(i+1<row)
                check(vec,i+1,j,row,col);
            if(j+1<col)
                check(vec,i,j+1,row,col);
        }
    }
};