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
SOLUTION 1:
经典 BFS 题目。如果使用DFS，会超时。
使用队列进行BFS。注意，在判断index越界时，主页君写了2种方法。
（1）可以直接对index在0-x*y之间取。这里有个小的trick，在点在左边缘的时候，index-1会导致我们回到上一行的最右边。当然那个点也是
边缘点，所以不会造成解的错误。
（2）判断点是不是在边缘，判定上下左右还有没有节点。
常出错的点：计算index是i * cols + j 这点要记好了。
public class Solution {
public void solve(char[][] board) {
if (board == null || board.length == 0 || board[0].length == 0) {
return;
}
int rows = board.length;
int cols = board[0].length;
// the first line and the last line.
for (int j = 0; j < cols; j++) {
bfs(board, 0, j);
bfs(board, rows - 1, j);
}
// the left and right column
for (int i = 0; i < rows; i++) {
bfs(board, i, 0);
bfs(board, i, cols - 1);
}
// capture all the nodes.
for (int i = 0; i < rows; i++) {
for (int j = 0; j < cols; j++) {
if (board[i][j] == 'O') {
board[i][j] = 'X';
} else if (board[i][j] == 'B') {
board[i][j] = 'O';
}
}
}
return;
}
public void bfs1(char[][] board, int i, int j) {
int rows = board.length;
int cols = board[0].length;
Queue<Integer> q = new LinkedList<Integer>();
q.offer(i * cols + j);
while (!q.isEmpty()) {
int index = q.poll();
// Index is out of bound.
if (index < 0 || index >= rows * cols) {
continue;
}
int x = index / cols;
int y = index % cols;
if (board[x][y] != 'O') {
continue;
}
board[x][y] = 'B';
q.offer(index + 1);
q.offer(index - 1);
q.offer(index + cols);
q.offer(index - cols);
}
}
public void bfs(char[][] board, int i, int j) {
int rows = board.length;
int cols = board[0].length;
Queue<Integer> q = new LinkedList<Integer>();
q.offer(i * cols + j);
while (!q.isEmpty()) {
int index = q.poll();
int x = index / cols;
int y = index % cols;
if (board[x][y] != 'O') {
continue;
}
board[x][y] = 'B';
if (y < cols - 1) {
q.offer(index + 1);
}
if (y > 0) {
q.offer(index - 1);
}
if (x > 0) {
q.offer(index - cols);
}
if (x < rows - 1) {
q.offer(index + cols);
}
}
}
}
SOLUTION 2:
附上DFS解法：
public void dfs(char[][] board, int i, int j) {
int rows = board.length;
int cols = board[0].length;
// out of bound or visited.
if (i < 0 || i >= rows || j < 0 || j >= cols) {
return;
}
if (board[i][j] != 'O') {
return;
}
board[i][j] = 'B';
// dfs the sorrounded regions.
dfs(board, i + 1, j);
dfs(board, i - 1, j);
dfs(board, i, j + 1);
dfs(board, i, j - 1);
}
GITHUB:
https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/bfs/Solve.java