[LeetCode]130. Surrounded Regions

Problem Description

[]https://leetcode.com/problems/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.

思路

就是围棋啊~~每个子的气是和周围子的气相同的。如果周围有活棋，该子就是活的。。。

自从不在自己的IDE测试后，AC率大大下降啊不开心！！！

Code

package q130;

public class Solution {
public int[][] helper(char[][] board) {
int m = board.length;
int n = board[0].length;
int[][] qi = new int[m + 1][n + 1];
for (int i = 0; i <= m; i++) {
for (int j = 0; j <= n; j++) {
qi[i][j] = 0;
}
}

for (int i = 0; i < m; i++) {
qi[i][n - 1] = board[i][n - 1] == 'X' ? 0 : 1;
qi[i][0] = board[i][0] == 'X' ? 0 : 1;
}
for (int i = 0; i < n; i++) {
qi[m - 1][i] = board[m - 1][i] == 'X' ? 0 : 1;
qi[0][i] = board[0][i] == 'X' ? 0 : 1;
}
System.out.println();
for (int k = 0; k < m; k++) {
for (int i = 1; i < m - 1; i++) {
for (int j = 1; j < n - 1; j++) {
qi[i][j] = (board[i][j] == 'X') ? 0 : qi[i - 1][j]
| qi[i][j - 1] | qi[i][j + 1] | qi[i + 1][j];
}
}
}

return qi;
}

public void solve(char[][] board) {
int m, n;
if (board == null || board.length == 0 || board[0].length == 0)
return;
m = board.length;
n = board[0].length;
int[][] qi = helper(board);
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (qi[i][j] == 0)
board[i][j] = 'X';
}
}

}
//
// public static void main(String[] args) {
// Solution s = new Solution();
// char[][] c = { "OXOOOX".toCharArray(), "OOXXXO".toCharArray(),
// "XXXXXO".toCharArray(), "OOOOXX".toCharArray(),
// "XXOOXO".toCharArray(), "OOXXXX".toCharArray() };
// int[][] p = s.helper(c);
// for (int i = 0; i < c.length; i++) {
// for (int j = 0; j < c[0].length; j++) {
// System.out.print(c[i][j] + " ");
// }
// System.out.println();
//
// }
// System.out.println();
// for (int i = 0; i < c.length; i++) {
// for (int j = 0; j < c[0].length; j++) {
// System.out.print(p[i][j] + " ");
// }
// System.out.println();
//
// }
// System.out.println();
//
// }

}