LeetCode----surrounded-regions

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本文介绍了一种解决二维棋盘中被'X'包围的所有'O'区域问题的算法。通过从棋盘边缘的'O'进行深度优先搜索，并标记可到达的'O'，最终将未被标记的'O'转化为'X'，实现'O'区域的捕获。

题目描述

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

思路：1,从四周的O开始深度搜索，将能搜到的O全置为×;

2,遍历所有元素，若当前元素为×将其置为O，若为O将其置为X。

代码实现：

class Solution {
public:
    void solve(vector<vector<char>> &board) {
        //从四周的O开始深度搜索，将能搜到的O全置为×
        //扫描所有元素，将×置为O，将O置为X。
        if (board.size() <= 2 || board[0].size() <=2)
            return ;
        for (int i=0; i<board[0].size(); ++i){
            dfs(board, 0, i);
            dfs(board, board.size()-1, i);
        }
        for (int i=1; i<board.size()-1; i++){
            dfs(board, i, 0);
            dfs(board, i, board[0].size()-1);
        }
        
        for (int i=0; i<board.size(); ++i)
            for (int j=0; j<board[0].size(); ++j){
            if (board[i][j] == '*')
                board[i][j] = 'O';
            else if (board[i][j] == 'O')
                board[i][j] = 'X';
        }
        return ;
    }
    
    //深度搜索
    void dfs(vector<vector<char>> &board, int row, int col){
        if (board[row][col] == 'O'){
            board[row][col] = '*';
            //
            if (row > 1)
                dfs(board, row-1, col);
            if (col > 1)
                dfs(board, row, col-1);
            if (row < board.size()-1){
                dfs(board, row+1, col);
            }
            if (col < board[0].size()-1){
                dfs(board, row, col+1);
            }
        }
        return ;
    }
};