[leetcode] 130. 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.

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.

分析

题目的意思是：如果O四周都是X的话，那么用X覆盖O，求最终的状态。

• 遍历矩阵的四条边，如果出现O字符，我们把O替换为Z，然后沿着这个点深度优先遍历找与之相邻的O，同样置为Z，最后我们把Z的位置用O替换，把原来的O用X替换

代码

class Solution {
public:
    void solve(vector<vector<char>>& board) {
        if(board.empty()||board.size()==0){
            return ;
        }
        int m=board.size();
        int n=board[0].size();
        
        for(int i=0;i<m;i++){
            if(board[i][0]=='O'){
                board[i][0]='Z';
                dfs(board,i,0,m,n);
            }
            if(board[i][n-1]=='O'){
                board[i][n-1]='Z';
                dfs(board,i,n-1,m,n);
            }
        }
        for(int i=0;i<n;i++){
            if(board[0][i]=='O'){
                board[0][i]='Z';
                dfs(board,0,i,m,n);
            }
            if(board[m-1][i]=='O'){
                board[m-1][i]='Z';
                dfs(board,m-1,i,m,n);
            }
        }
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(board[i][j]=='O'){
                    board[i][j]='X';
                }
                if(board[i][j]=='Z'){
                    board[i][j]='O';
                }
            }
        }
    }
    void dfs(vector<vector<char>>& board,int x,int y,int m,int n){
        int dx[]={-1,0,0,1};
        int dy[]={0,-1,1,0};
        for(int i=0;i<4;i++){
            int x1=x+dx[i];
            int y1=y+dy[i];
            if(!isOut(x1,y1,m,n)&&board[x1][y1]=='O'){
                board[x1][y1]='Z';
                dfs(board,x1,y1,m,n);
            }
        }
    }
    bool isOut(int x,int y,int m,int n){
        return x<0||x>=m||y<0||y>=n;
    }
};

参考文献

​​Leetcode—Surrounded Regions​​​​[编程题]surrounded-regions​​