搜索（DFS）---填充封闭区域

填充封闭区域

130. Surrounded Regions (Medium)

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

题目描述：

  使被'X'包围的‘O’转换为'X'。

思路分析：

  我们先将与边界上所有O相联通的O标记，然后遍历数组，遇见未被标记的O就将其换成X，这样就能保证所有被X包围的O都被替换成X。

代码：

class Solution {
    public void solve(char [][]board){
    if(board==null||board.length==0)
        return ;
    int m=board.length;
    int n=board[0].length;
    for(int i=0;i<n;i++){
        if(board[0][i]=='O'){ //上边界
            dfs(board,0,i);
        }
        if(board[m-1][i]=='O'){ //下边界
            dfs(board,m-1,i);
        }
    }
    for(int j=0;j<m;j++){
        if(board[j][0]=='O')
            dfs(board,j,0);
        if(board[j][n-1]=='O')
            dfs(board,j,n-1);
    }
    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]=='0')//与边界联通的O
                board[i][j]='O';
        }
    }
     return;   
}
public void dfs(char[][]board,int i,int j){ //标记与边界O相连的O
    if(board[i][j]!='O')
        return ;
    board[i][j]='0';
    if(i+1<board.length)
        dfs(board,i+1,j);
    if(i-1>=0)
        dfs(board,i-1,j);
    if(j+1<board[0].length)
        dfs(board,i,j+1);
    if(j-1>=0)
        dfs(board,i,j-1);
}
}