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

 

思路一：深度优先+Flag

深度优先 DFS 遍历矩阵的边缘元素，将O相连的区域赋值为1，其他的为0，在将flag标志为0的board置为X。

实现代码

class Solution {     
public:
    void solve(vector> &board) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int rows = board.size();
        if(rows<=0)
            return;
        int cols = board[0].size();
        vector> flag(rows,vector(cols,0));          
        for(int i=0;i> &board,vector> &flag)
    {
        int rows = board.size();
        int cols = board[0].size();
        if(i<0||i>=rows||j<0||j>=cols)
            return ;
        if(board[i][j]=='X')
            return ;  
        if(board[i][j]=='O' && flag[i][j]==0)
        {
            flag[i][j]=1;
            dfs(i,j-1,board,flag);
            dfs(i,j+1,board,flag);
            dfs(i+1,j,board,flag);
            dfs(i-1,j,board,flag);
        }
         
    }
};

注意：要对 flag 数组初始化 vector> flag(rows,vector(cols,0));  否则会出现错误。

上述运行结果是 对于 Juge Small 运行正确，但是对于 Juge Large 会出现 Runtime Error 。

因该是 运行时内存溢出，主要是用到了 flag 数组。其实完全不必用到的。下面的改进。

思路二：深度优先 

 1、First scan the four edges of the board, if you meet an 'O', call a recursive mark function to mark that region to something else (for example, '+');

2、scan all the board, if you meet an 'O', flip it to 'X';

3、scan again the board, if you meet an '+', flip it to 'O';

step 2 and 3 can be merged.

然后我实现的代码如下：

class Solution {         
public:    
    void dfs(int i,int j,vector> &board)    
    { 
        if(i<0 || i>=board.size() || j<0 || j>=board[0].size() )  
            return;  
        if(board[i][j]=='O')    
        {    
            board[i][j]='#';  
            dfs(i,j+1,board);  
            dfs(i,j-1,board);  
            dfs(i-1,j,board);   
            dfs(i+1,j,board);   
        }    
    }  
    void solve(vector> &board) {    
        // Start typing your C/C++ solution below    
        // DO NOT write int main() function    
        int rows = board.size();    
        if(rows<=2)  
            return ;  
        int cols = board[0].size();            
        for(int i=0;i

这样就可以完全的 Juge Large 了。 

思路三：广度优先BFS

class Solution {
private:
    struct Node
    {
        int x,y;
        Node(){};
        Node(int a,int b):x(a),y(b){};
    };
public:
    void solve(vector> &board) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int rows = board.size();
        if(rows<=2) return ;
        int cols = board[0].size();
        queue myqueue;
        for(int i=0;i=board.size() || ty<0 || ty>=board[0].size() || board[tx][ty]!='O')
                    continue ;
                board[tx][ty]='#';
                myqueue.push(Node(tx,ty));
            }
            
        }
        
        for(int i=0;i

当然可以不用建立 Node 结构体，而是  queue> Q;      Q.push(make_pair(i , j)); 来实现。

速度更快。

最新版本：

public class Solution {
    
    // use a queue to do BFS
	private Queue queue = new LinkedList();
    
    public void solve(char[][] board) {
        if(board == null){
            return;
        }
        int row = board.length;
        if(row <= 2) {
            return;
        }
        int col = board[0].length;
        if(col <= 2){
            return;
        }
        
        for(int i=0; i= row || j < 0 || j>=col){
            return;
        }
        if(board[i][j] == 'O'){
            board[i][j] = '#';
            queue.offer(i*col+j);
        }
    }
    
    
    // DFS: Line 50: java.lang.StackOverflowError
    private void dfs(int i, int j, char[][] board) {
        int row = board.length;
        int col = board[0].length;
        if(i < 0 || i >= row || j < 0 || j>=col){
            return;
        }
        if(board[i][j] == 'O'){
            board[i][j] = '#';
            dfs(i+1, j, board);
            dfs(i-1, j, board);
            dfs(i, j+1, board);
            dfs(i, j-1, board);
        }
        
    }
    
}