腾讯笔经面经（微信事业群）
文章目录
- 算法题
- 1、题目1
- 2、题目2
- 3、题目三
- 技术面试
- 疑问解答
微信事业群
时间：2019/9/10 11:00 ～ 13:30
算法题
1、题目1
对于一棵满二叉排序树深度为K，节点数为 2^K - 1 ；节点值为 1至 (2^K-1)。
给出K和任意三个节点的值，输出包含该三个节点的最小子树的根节点值
样例输入：4 10 15 13
样例输出：12
我的代码实现：
package NormalTest;
import java.util.Scanner;
public class Main1 {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
while (in.hasNextInt()) {
String dataStr = in.nextLine();
String[] strs = dataStr.split(" ");
int dep = Integer.parseInt(strs[0]);
int[] array = new int[dep-1];
for (int i = 0; i < dep-1; i++) {
array[i] = Integer.parseInt(strs[i+1]);
}
System.out.println(get(dep, array));
}
}
private static int get (int dep, int[] array) {
int cur = (int)Math.pow(2,dep-1);
int result = 0;
for (int i = dep-1; i > 0; i--) {
if (array[0] > cur && array[1] > cur && array[2] > cur)
cur = cur + (int)Math.pow(2, i-1);
else if (array[0] < cur && array[1] < cur && array[2] < cur)
cur = cur - (int)Math.pow(2, i-1);
else
{
result = cur;
break;
}
}
return result;
}
}
2、题目2
回形矩阵是由1开始的自然数顺时针排列成的一个n*n矩阵,n为奇数.
1 2 3 4 5
16 17 18 19 6
15 24 25 20 7
14 23 22 21 8
13 12 11 10 9
要求打印出它的一个子矩阵(m*m),例如在n=5的矩阵里面以起始点 xy( 2,2 ) 打印m = 2的正方形，则输出:
17,18
24,25
进阶： 是否可以应对超大规模的情况，例如n > 10000000,m<100的场景
我的代码实现
package NormalTest;
import java.util.Scanner;
public class Main2 {
//private static int length = 0;
private static int value = 1;
//private static int[][] array = null;
private static Direction lastDirection = Direction.Right;
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
while (in.hasNext()) {
String dataStr = in.nextLine();
String[] strs = dataStr.split(" ");
int length = Integer.parseInt(strs[0]);
int cow = Integer.parseInt(strs[1]);
int col = Integer.parseInt(strs[2]);
int[][] array = new int[length][length];
init(length, array);
print(array, length, cow, col);
}
}
private static enum Direction {
Right, Down, Left, Up;
}
public static void init(int length, int[][] array) {
int row = 0, col = 0;
for (int i = 0; i < length * length; i++) {
array[row][col] = value;
lastDirection = findDirection(row, col, length);
switch (lastDirection) {
case Right:
col++;
break;
case Down:
row++;
break;
case Left:
col--;
break;
case Up:
row--;
break;
default:
System.out.println("error");
}
value++;
}
}
private static Direction findDirection(int row, int col, int length) {
Direction direction = lastDirection;
switch (direction) {
case Right: {
if (row == 0 && col >= length-1)
direction=Direction.Down;
break;
}
case Down: {
if (row == length-1 && col >= length-1)
direction=Direction.Left;
break;
}
case Left: {
if (row == length-1&&col <= 0)
direction=Direction.Up;
break;
}
case Up: {
if (row == 1 && col <= 0)
direction=Direction.Right;
break;
}
}
return direction;
}
private static void print(int[][] arr, int length) {
for (int i = 0; i < length; i++) {
for (int j = 0; j < length; j++) {
System.out.printf(" %2d", arr[i][j]);
}
System.out.println();
}
}
private static void print(int[][] arr, int length, int cow, int col) {
for (int i = cow-1; i < length; i++) {
for (int j = col-1; j < length; j++) {
System.out.printf(" %2d", arr[i][j]);
}
System.out.println();
}
}
}
3、题目三
一个页面有两个广告位，现有四个广告需要展示，实现一个算法输出需要展示的两个广告
要求：
a.输出的两个广告不能相同；
b.多次调用输出的四个广告的概率为：1:2:3:4 （4个广告用整形1-4来表示）
c.进阶：是否存在通用解，n个广告，概率为p1:p2:…:pn
我的代码实现：
import java.util.Scanner;
public class Main2 {
//private static int length = 0;
private static int value = 1;
//private static int[][] array = null;
private static Direction lastDirection = Direction.Right;
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
while (in.hasNext()) {
String dataStr = in.nextLine();
String[] strs = dataStr.split(" ");
int length = Integer.parseInt(strs[0]);
int cow = Integer.parseInt(strs[1]);
int col = Integer.parseInt(strs[2]);
int[][] array = new int[length][length];
init(length, array);
print(array, length, cow, col);
}
}
private static enum Direction {
Right, Down, Left, Up;
}
public static void init(int length, int[][] array) {
int row = 0, col = 0;
for (int i = 0; i < length * length; i++) {
array[row][col] = value;
lastDirection = findDirection(row, col, length);
switch (lastDirection) {
case Right:
col++;
break;
case Down:
row++;
break;
case Left:
col--;
break;
case Up:
row--;
break;
default:
System.out.println("error");
}
value++;
}
}
private static Direction findDirection(int row, int col, int length) {
Direction direction = lastDirection;
switch (direction) {
case Right: {
if (row == 0 && col >= length-1)
direction=Direction.Down;
break;
}
case Down: {
if (row == length-1 && col >= length-1)
direction=Direction.Left;
break;
}
case Left: {
if (row == length-1&&col <= 0)
direction=Direction.Up;
break;
}
case Up: {
if (row == 1 && col <= 0)
direction=Direction.Right;
break;
}
}
return direction;
}
private static void print(int[][] arr, int length) {
for (int i = 0; i < length; i++) {
for (int j = 0; j < length; j++) {
System.out.printf(" %2d", arr[i][j]);
}
System.out.println();
}
}
private static void print(int[][] arr, int length, int cow, int col) {
for (int i = cow-1; i < length; i++) {
for (int j = col-1; j < length; j++) {
System.out.printf(" %2d", arr[i][j]);
}
System.out.println();
}
}
}
技术面试
说说项目
- 挑战？
- 如何解决痛点问题的
- 分布式事务
谈到2PC、3阶段，TCC、AT、Saga、TX
- 空回滚是什么？事务悬挂又是什么？
事务悬挂是一阶段缺失二阶段的现象
事务空回滚是因为一阶段由于网络阻塞导致超时二阶段回滚，之后一阶段恢复之后导致事务悬挂
- 事务悬挂如何解决？
5.1 及时发现，预警
5.2 30秒/60秒拒绝空回滚
事务操作过程记录事务的状态，有空回滚的事务及时处理
疑问解答
博主实时在线，欢迎读者交流，相互学习