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130.surrounded-regions.java
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130.surrounded-regions.java
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/*
* @lc app=leetcode id=130 lang=java
*
* [130] Surrounded Regions
*
* https://leetcode.com/problems/surrounded-regions/description/
*
* algorithms
* Medium (28.67%)
* Likes: 2422
* Dislikes: 734
* Total Accepted: 277.2K
* Total Submissions: 950.4K
* Testcase Example: '[["X","X","X","X"],["X","O","O","X"],["X","X","O","X"],["X","O","X","X"]]'
*
* Given a 2D board containing 'X' and 'O' (the letter O), capture all regions
* surrounded by 'X'.
*
* A region is captured by flipping all 'O's into 'X's in that surrounded
* region.
*
* 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
*
*
* Explanation:
*
* Surrounded regions shouldn’t be on the border, which means that any 'O' on
* the border of the board are not flipped to 'X'. Any 'O' that is not on the
* border and it is not connected to an 'O' on the border will be flipped to
* 'X'. Two cells are connected if they are adjacent cells connected
* horizontally or vertically.
*
*/
// @lc code=start
class Solution {
public void solve(char[][] board) {
if (board.length == 0 || board[0].length == 0) return;
if (board.length < 3 || board[0].length < 3) return;
int m = board.length;
int n = board[0].length;
for (int i = 0; i < m; i++) {
if (board[i][0] == 'O') helper(board, i, 0);
if (board[i][n - 1] == 'O') helper(board, i, n - 1);
}
for (int j = 1; j < n - 1; j++) {
if (board[0][j] == 'O') helper(board, 0, j);
if (board[m - 1][j] == 'O') helper(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';
}
}
}
private void helper(char[][] board, int r, int c) {
if (r < 0 || c < 0 || r > board.length - 1 || c > board[0].length - 1 || board[r][c] != 'O') return;
board[r][c] = '*';
helper(board, r + 1, c);
helper(board, r - 1, c);
helper(board, r, c + 1);
helper(board, r, c - 1);
}
}
// @lc code=end