被包围的区域 Surrounded Regions

有一个 m*n的矩阵，每个坐标都有字符X ,O组成，找到所有被X包围的O区域。将O换成X

in：[["X","X","X","X"],["X","O","O","X"],["X","X","O","X"],["X","O","X","X"]]
out：[["X","X","X","X"],["X","X","X","X"],["X","X","X","X"],["X","O","X","X"]]

思路

使用DFS，借助方向数组处理,从4边开始将所有与边上的O相连的O，都标记成其他字符例如#。

即所有#都不能修改成X，从一开始遍历将所有的O都改成X，遇到#改成O。

public int n;//column
public int m;//row
public int[][] dir ={{0,1},{1,0},{0,-1},{-1,0}};
public void solve(char[][] board) {
	m = board.length;// row
	n = board[0].length;//column
	
	for(int i =0;i<m;i++){
		if(board[i][0]=='O'){
			dfs(i,0,board);
		}			
		if(board[i][n-1]=='O'){
			dfs(i,n-1,board);
		}
	}
	for(int j =0;j<n;j++){
		if(board[0][j]=='O'){
			dfs(0,j,board);
		}
		if(board[m-1][j]=='O'){
			dfs(m-1,j,board);
		}
	}
	
	for(int i=0;i<m;i++){
		for(int j=0;j<n;j++){
			if(board[i][j]=='O'){
				board[i][j]='X';
			}
			else if(board[i][j]=='#'){
				board[i][j]='O';
			}
		}
	}
}

public void dfs(int i,int j,char[][] arr){
	arr[i][j]='#';
	for(int k =0;k<4;k++){
		int x = i+dir[k][0];
		int y = j+dir[k][1];
		if(x<0||x>=m||y<0||y>=n){
			continue;
		}
		if(arr[x][y]!='O'){
			continue;
		}
		dfs(x,y,arr);
	}
	return;
}

Tag

DFS