用代码实现给定一个n*n的包含‘X’和‘O’的面板,捕捉所有被‘X’环绕的区域,并将该区域中的所有‘O’翻转为‘X’
时间: 2023-09-26 10:05:49 浏览: 55
好的,以下是实现这个功能的Java代码:
```
public void solve(char[][] board) {
if (board == null || board.length == 0 || board[0].length == 0) {
return;
}
int m = board.length;
int n = board[0].length;
for (int i = 0; i < m; i++) {
if (board[i][0] == 'O') {
dfs(board, i, 0);
}
if (board[i][n - 1] == 'O') {
dfs(board, i, n - 1);
}
}
for (int j = 0; j < n; j++) {
if (board[0][j] == 'O') {
dfs(board, 0, j);
}
if (board[m - 1][j] == 'O') {
dfs(board, m - 1, j);
}
}
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] == '#') {
board[i][j] = 'O';
}
}
}
}
public void dfs(char[][] board, int i, int j) {
int m = board.length;
int n = board[0].length;
if (i < 0 || i >= m || j < 0 || j >= n || board[i][j] == 'X' || board[i][j] == '#') {
return;
}
board[i][j] = '#';
dfs(board, i - 1, j);
dfs(board, i + 1, j);
dfs(board, i, j - 1);
dfs(board, i, j + 1);
}
```
该代码采用了深度优先搜索的方法,首先遍历边界上的'O',将与边界上'O'相邻的'O'都标记为'#',最后将所有剩下的'O'翻转为'X'即可。算法的时间复杂度为$O(n^2)$,空间复杂度为$O(n^2)$。