Description
Given an m x n
matrix board
containing 'X'
and 'O'
, capture all regions that are 4-directionally surrounded by 'X'
.
A region is captured by flipping all 'O'
s into 'X'
s in that surrounded region.
Example 1:
Input: board = [[“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”]] Explanation: Notice that an ‘O’ should not be flipped if:
- It is on the border, or
- It is adjacent to an ‘O’ that should not be flipped. The bottom ‘O’ is on the border, so it is not flipped. The other three ‘O’ form a surrounded region, so they are flipped.
Example 2:
Input: board = “X” Output: “X”
Constraints:
m == board.length
n == board[i].length
1 <= m, n <= 200
board[i][j]
is'X'
or'O'
.
Code
Boundary DFS,從邊界開始著手,由外而內。
class Solution {
public:
void solve(vector<vector<char>>& board) {
int m = board.size();
int n = board[0].size();
vector<vector<int>> visited(m, vector<int>(n, 0));
for(int i = 0; i < m; i++) {
if(board[i][0] == 'O') {
dfs(i, 0, m, n, board);
}
if(board[i][n-1] == 'O') {
dfs(i, n-1, m, n, board);
}
}
for(int i = 0; i < n; i++) {
if(board[0][i] == 'O') {
dfs(0, i, m, n, board);
}
if(board[m-1][i] == 'O') {
dfs(m-1, i, m, n, board);
}
}
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] == 'Z') {
board[i][j] = 'O';
}
}
}
}
void dfs(int x, int y, int m, int n, vector<vector<char>>& board) {
if(x >= m || y >= n || x < 0 || y < 0 || board[x][y] != 'O') return;
board[x][y] = 'Z';
dfs(x+1, y, m, n, board);
dfs(x-1, y, m, n, board);
dfs(x, y+1, m, n, board);
dfs(x, y-1, m, n, board);
}
};