题解：
- 枚举二维数组每个位置，向上、下、左、右四个方向能延伸的最长长度
- 取这四个方向的最小值，即为答案
- 可以用f[i] = f[i - 1] + 1 计算四个方向的最大值，然后参考滚动数组思想，用一个变量记录即可
class Solution {
public int orderOfLargestPlusSign(int n, int[][] mines) {
int[][] f = new int[510][];
boolean[][] g = new boolean[510][];
for (int i = 0; i < n; i++) {
g[i] = new boolean[510];
for (int j = 0; j < n; j++) {
g[i][j] = true;
}
}
for (int[] mine : mines) {
g[mine[0]][mine[1]] = false;
}
for (int i = 0; i < n; i++) {
f[i] = new int[510];
int s = 0;
for (int j = 0; j < n; j++) {
if (g[i][j]) s++;
else s = 0;
f[i][j] = s;
}
}
for (int i = 0; i < n; i++) {
int s = 0;
for (int j = n - 1; j >= 0; j--) {
if (g[i][j]) s++;
else s = 0;
f[i][j] = Math.min(f[i][j], s);
}
}
for (int i = 0; i < n; i++) {
int s = 0;
for (int j = 0; j < n; j++) {
if (g[j][i]) s++;
else s = 0;
f[j][i] = Math.min(f[j][i], s);
}
}
for (int i = 0; i < n; i++) {
int s = 0;
for (int j = n - 1; j >= 0; j--) {
if (g[j][i]) s++;
else s = 0;
f[j][i] = Math.min(f[j][i], s);
}
}
int res = 0;
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
res = Math.max(res, f[i][j]);
return res;
}
}