[leetcode] 130.Surrounded Regions

题目：
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
题意：
给一个二维的表格，里面包含了’X’和’O’，找出所有被X包围的范围，将里面的O改成O。
思路：
这个题目有点像下围棋的感觉，我们不妨可以反过来思考这个问题。题目让找出被X包围的O并将其改为X。我们不妨找出那些不能被修改的O，不能被包围的部分是与四个边上的O相邻的O，相邻的意思是这两个点的位置处于上下位置或者左右位置。如果不能从四个边到达的O一定是被X包围起来的，所以我们的目标就变成了找出所有与四个边上的O相邻的所有 O，然后把它们标记为‘+’这类符号，把剩下的O全部改成X即可。
代码如下：
采用广度优先搜索，把所有的四个边上的O的位置找出来，并把这些O改为+，接下来通过广度优先搜索的方法找出其他的O。

class Solution {public:    void solve(vector<vector<char>>& board) {        int row = board.size();        if(row == 0)return;        int col = board[0].size();        if(col == 0)return;        queue<pair<int,int>> OIndex;        for(auto i = 0; i < col; i++) {            if(board[0][i] == 'O') {                OIndex.push(make_pair(0, i));                board[0][i] = '+';            }            if(board[row - 1][i] == 'O') {                OIndex.push(make_pair(row-1, i));                board[row - 1][i] = '+';            }        }        for(auto i = 0; i < row; i++) {            if(board[i][0] == 'O') {                OIndex.push(make_pair(i, 0));                board[i][0] = '+';            }            if(board[i][col - 1] == 'O') {                OIndex.push(make_pair(i, col - 1));                board[i][col - 1] = '+';            }        }        while(!OIndex.empty()) {            pair<int,int> index = OIndex.front();            OIndex.pop();            if(index.first != 0 && board[index.first - 1][index.second] == 'O') {                board[index.first - 1][index.second] = '+';                OIndex.push(make_pair(index.first - 1, index.second));            }            if(index.first != row - 1 && board[index.first + 1][index.second] == 'O') {                board[index.first + 1][index.second] = '+';                OIndex.push(make_pair(index.first + 1, index.second));            }            if(index.second != 0 && board[index.first][index.second - 1] == 'O') {                board[index.first][index.second - 1] = '+';                OIndex.push(make_pair(index.first, index.second - 1));            }            if(index.second != col - 1 && board[index.first][index.second + 1] == 'O') {                board[index.first][index.second + 1] = '+';                OIndex.push(make_pair(index.first, index.second + 1));            }        }        for(auto i = 0; i < row; i++)            for(auto j = 0; j < col; j++){                if(board[i][j] == 'O')board[i][j] = 'X';                else if(board[i][j] == '+')board[i][j] = 'O';            }    }};