8.27 网易互娱笔试
2个半小时 3道编程题
第一题 铺墙纸 简单找找规律挺简单，只不过处理数据需要细心耐心一点
第二题 找有效矩形面积 判断相交然后计算矩形面积然后减去重合的面积即可（难度不大，但是自己还是太菜了，最后才发现要减去重合面积，已经来不及了....）
第三题 手势密码 完全没思路没时间
有心理准备网易的笔试会很难，准备了很久很久还是不如人意。无论什么原因都只是借口，只能怪自己能力还是太差了。
贴一下第一题ac代码
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int T = Integer.parseInt(sc.nextLine());
List<char[][]> data1 = new ArrayList<>();
int[][] data2 = new int[T][2];
for (int i = 0; i < T; i++) {
String[] info = sc.nextLine().split(" ");
int n = Integer.parseInt(info[0]);
int m = Integer.parseInt(info[1]);
char[][] room = new char[n][n];
for (int j = 0; j < n; j++) {
String rowStr = sc.nextLine();
char[] row = rowStr.toCharArray();
for (int k = 0; k < row.length; k++) {
room[j][k] = row[k];
}
}
data1.add(room);
data2[i][0] = n;
data2[i][1] = m;
if (i != T - 1) {
sc.nextLine();
}
}
for (int i = 0; i < data1.size(); i++) {
getResult(data1.get(i), data2[i][0], data2[i][1]);
}
}
public static void getResult(char[][] room, int n, int m) {
// 求差值 也就是扩容的次数
int diff = (m - n) / 2;
// 扩容后的数组
char[][] result = new char[m][m];
int leftBoundary = diff;
int rightBoundary = diff + n - 1;
int topBoundary = diff;
int bottomBoundary = diff + n - 1;
// 把原数组填进去
for (int i = topBoundary; i <= bottomBoundary; i++) {
for (int j = leftBoundary; j <= rightBoundary; j++) {
result[i][j] = room[i - diff][j - diff];
}
}
for (int i = 0; i < diff; i++) {
// 左右扩充
for (int j = topBoundary; j <= bottomBoundary; j++) {
result[j][leftBoundary - 1] = result[j][leftBoundary + (2 * i)];
result[j][rightBoundary + 1] = result[j][rightBoundary - (2 * i)];
}
leftBoundary--;
rightBoundary++;
// 上下扩充
for (int j = leftBoundary; j <= rightBoundary; j++) {
result[topBoundary - 1][j] = result[topBoundary + (2 * i)][j];
result[bottomBoundary + 1][j] = result[bottomBoundary - (2 * i)][j];
}
topBoundary--;
bottomBoundary++;
}
for (char[] chars : result) {
for (char aChar : chars) {
System.out.print(aChar);
}
System.out.println();
}
System.out.println();
}
}
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