标签：empty oid 节点 script for 一个 out nbsp int
‘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.
Example 1:
Input:
X X X X
X O O X
X X O X
X O X X
Output:
X X X X
X X X X
X X X X
X O X X
Example 2:
Input:
X X X X
X O O X
X O O X
X O X X
Output:
X X X X
X O O X
X O O X
X O X X
思路：可以使用BFS或DFS解题.
方法1:
在记录每个节点是否访问过的前提下, 依次从每个 ‘O‘ 开始BFS/DFS, 并且只访问未访问过的 ‘O‘.
如果从一个 ‘O‘ 可以访问到边界, 那么不做任何操作; 否则便将这个 ‘O‘ 可以访问到的所有的 ‘O‘ 替换为 ‘X‘.
方法2:
从每个边界的 ‘O‘ 开始遍历, 只访问 ‘O‘, 先都暂时设置为 ‘T‘ 或其他字符.
遍历结束之后, 将剩下的 ‘O‘ 替换为 ‘X‘ 然后再将 ‘T‘ 还原即可.
public class Solution {
public void surroundedRegions(char[][] board) {
// Write your code here
int n = board.length;
if (n == 0) {
return;
}
int m = board[0].length;
for (int i = 0; i < n; i++) {
bfs(board, i, 0);
bfs(board, i, m - 1);
}
for (int j = 0; j < m; j++) {
bfs(board, 0, j);
bfs(board, n - 1, j);
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (board[i][j] == ‘W‘) {
board[i][j] = ‘O‘;
} else {
board[i][j] = ‘X‘;
}
}
}
}
void bfs(char[][] board, int sx, int sy) {
if (board[sx][sy] != ‘O‘) {
return;
}
int n = board.length;
int m = board[0].length;
int[] dx = { 0, 1, 0, -1 };
int[] dy = { 1, 0, -1, 0 };
Queue<Integer> qx = new LinkedList<>();
Queue<Integer> qy = new LinkedList<>();
qx.offer(sx);
qy.offer(sy);
board[sx][sy] = ‘W‘; // ‘W‘ -> Water
while (!qx.isEmpty()) {
int cx = qx.poll();
int cy = qy.poll();
for (int i = 0; i < 4; i++) {
int nx = cx + dx[i];
int ny = cy + dy[i];
if (0 <= nx && nx < n && 0 <= ny && ny < m && board[nx][ny] == ‘O‘) {
board[nx][ny] = ‘W‘; // ‘W‘ -> Water
qx.offer(nx);
qy.offer(ny);
}
}
}
}
}
标签：empty oid 节点 script for 一个 out nbsp int
原文地址：https://www.cnblogs.com/FLAGyuri/p/12078591.html
