给定一个二维的矩阵,包含 'X'
和 'O'
(字母 O)。
找到所有被 'X'
围绕的区域,并将这些区域里所有的 'O'
用 'X'
填充。
示例:
X X X X X O O X X X O X X O X X
运行你的函数后,矩阵变为:
X X X X X X X X X X X X X O X X
解释:
被围绕的区间不会存在于边界上,换句话说,任何边界上的 'O'
都不会被填充为 'X'
。 任何不在边界上,或不与边界上的 'O'
相连的 'O'
最终都会被填充为 'X'
。如果两个元素在水平或垂直方向相邻,则称它们是“相连”的。
C
void DFS(char** board,int row,int col,int m,int n)
{
board[row][col]='a';
if(row-1>=0 && 'O'==board[row-1][col])
{
DFS(board,row-1,col,m,n);
}
if(row+1<=m-1 && 'O'==board[row+1][col])
{
DFS(board,row+1,col,m,n);
}
if(col-1>=0 && 'O'==board[row][col-1])
{
DFS(board,row,col-1,m,n);
}
if(col+1<=n-1 && 'O'==board[row][col+1])
{
DFS(board,row,col+1,m,n);
}
}
void solve(char** board, int boardRowSize, int boardColSize)
{
int m=boardRowSize;
int n=boardColSize;
if(0==m || 0==n)
{
return;
}
for(int i=0;i<n;i++)
{
if('O'==board[0][i])
{
DFS(board,0,i,m,n);
}
if('O'==board[m-1][i])
{
DFS(board,m-1,i,m,n);
}
}
for(int i=0;i<m;i++)
{
if('O'==board[i][0])
{
DFS(board,i,0,m,n);
}
if('O'==board[i][n-1])
{
DFS(board,i,n-1,m,n);
}
}
for(int i=0;i<m;i++)
{
for(int j=0;j<n;j++)
{
if('O'==board[i][j])
{
board[i][j]='X';
}
if('a'==board[i][j])
{
board[i][j]='O';
}
}
}
}
C++
class Solution {
public:
void DFS(vector<vector<char>>& board,int row,int col)
{
board[row][col]='a';
if(row-1>=0 && 'O'==board[row-1][col])
{
DFS(board,row-1,col);
}
if(row+1<=board.size()-1 && 'O'==board[row+1][col])
{
DFS(board,row+1,col);
}
if(col-1>=0 && 'O'==board[row][col-1])
{
DFS(board,row,col-1);
}
if(col+1<=board[0].size() && 'O'==board[row][col+1])
{
DFS(board,row,col+1);
}
}
void solve(vector<vector<char>>& board)
{
if(board.empty() || board[0].empty())
{
return;
}
int m=board.size();
int n=board[0].size();
for(int i=0;i<n;i++)
{
if('O'==board[0][i])
{
DFS(board,0,i);
}
if('O'==board[m-1][i])
{
DFS(board,m-1,i);
}
}
for(int i=0;i<m;i++)
{
if('O'==board[i][0])
{
DFS(board,i,0);
}
if('O'==board[i][n-1])
{
DFS(board,i,n-1);
}
}
for(int i=0;i<m;i++)
{
for(int j=0;j<n;j++)
{
if('O'==board[i][j])
{
board[i][j]='X';
}
if('a'==board[i][j])
{
board[i][j]='O';
}
}
}
}
};
python
class Solution:
def DFS(self,board,row,col):
board[row][col]='a'
if row-1>=0 and 'O'==board[row-1][col]:
self.DFS(board,row-1,col)
if row+1<=len(board)-1 and 'O'==board[row+1][col]:
self.DFS(board,row+1,col)
if col-1>=0 and 'O'==board[row][col-1]:
self.DFS(board,row,col-1)
if col+1<=len(board[0])-1 and 'O'==board[row][col+1]:
self.DFS(board,row,col+1)
def solve(self, board):
"""
:type board: List[List[str]]
:rtype: void Do not return anything, modify board in-place instead.
"""
if []==board or [[]]==board:
return
m=len(board)
n=len(board[0])
for i in range(n):
if 'O'==board[0][i]:
self.DFS(board,0,i)
if 'O'==board[m-1][i]:
self.DFS(board,m-1,i)
for i in range(m):
if 'O'==board[i][0]:
self.DFS(board,i,0)
if 'O'==board[i][n-1]:
self.DFS(board,i,n-1)
for i in range(m):
for j in range(n):
if 'O'==board[i][j]:
board[i][j]='X'
if 'a'==board[i][j]:
board[i][j]='O'