【leetcode题解】5 - Surrounded Regions

Surrounded Regions

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

leetcode题目链接

解题思路：

边缘的'O'一定不会被'X'包围，和边缘的'O'连通的'O'也不会被包围

用DFS BFS 搜索边缘'O'的联通'O'（上下左右四个方向），解决问题

源码：

class Solution {
public:
// 从该点开始在上下左右四个方向上做深度优先搜索
// 直到不是'O' 找到连通的'O'
void DFS(vector<vector<char> > &board, int row, int col)
{
if ((row < 0) || (col < 0) ||
(row >= board.size()) || (col >= board[0].size()) ||
(board[row][col] != 'O') )
return;

board[row][col] = '#';      // 连通的'O'标记为'#'
DFS(board, row - 1, col);   // up
DFS(board, row + 1, col);   // down
DFS(board, row, col - 1);   // right
DFS(board, row, col + 1);   // left

return;
}

void solve(vector<vector<char> > &board) {
// Note: The Solution object is instantiated only once and is reused by each test case.

// 边缘的'O'一定不会被'X'包围
// 和边缘的'O'连通的'O'也不会被包围
// 用DFS BFS进行搜索，解决问题

if (board.empty())
return;

int n = board.size();
int m = board[0].size();
for (int i = 0; i < n; ++i)  // row
{
DFS(board, i, 0);        // 第i行从第0列开始DFS
DFS(board, i, m-1);      // 第i行从第m-1列开始DFS
}

// for (int j = 0; j < m; ++j)  // col
for (int j = 1; j < m-1; ++j)  // 第一列和最后一列已经在上面遍历了
{
DFS(board, 0, j);        // 第0列
DFS(board, n-1, j);      // 第n-1列
}

for (int i = 0; i < n; ++i)
{
for (int j = 0; j < m; ++j)
{
if (board[i][j] == 'O')
board[i][j] = 'X';
else if (board[i][j] == '#')
board[i][j] = 'O';
}
}
return;
}
};

以上用DFS做的，大数据通不过。

试了网上其他人的代码，也都通不过。

法2：BFS

ing……