Analysis of Written Test Questions in Java Interview

• topic

写一个验证掷骰子概率的程序，同时投掷 2 颗 6 面骰子 n 次，计算其和得到 各数字的概率

• Analysis
  According to my own analysis steps, first analyze and calculate the sum of the numbers obtained by 2 6-sided dice. But how to calculate the sum? That is to generate a number between Random and 1-6 (new Random ().nextInt(6)+1) each time through a random number. This represents a dice. Each time you write the number that appears twice, it adds up to 2 to get the number.的和。 The sum. Then use the sum as the key of the Map, and then the Value is the number of times. The first is 1 later you can use the new Java8 Map method merge. I haven't used it here for the time being because it is easy to understand. The outside is n cycles. At this time, there are already n times of different values ​​in the map. The next step is to calculate the probability. After analysis, the denominator is n and the number of occurrences is the numerator. This player who often plays mahjong should know that both are 1 and 2 and both are 6 and 12. So no matter how many times they appear and the range is 2-12. So according to this cycle, the probability of each occurrence is calculated.

• Code

package com.dairuijie.demo.study;

import java.math.BigDecimal;
import java.util.HashMap;
import java.util.Map;
import java.util.Random;
import java.util.Scanner;

/**
 * 
 * @模块名：Day01
 * @包名：com.dairuijie.demo.study
 * @描述：SievesTest.java @版本：1.0
 * @创建人：drj
 * @创建时间：2020年3月28日下午12:35:14
 */
public class SievesTest {
    
    

	public static void main(String[] args) {
    
    
		Scanner scan = new Scanner(System.in);
		while (true) {
    
    
			try {
    
    
				System.out.print("请输入次数：");
				Integer read = Integer.valueOf(scan.nextLine());
				if (read == 0) {
    
    
					break;
				}
				System.out.println("输入数据：" + read);
				long s = System.currentTimeMillis();
				Random r = new Random();
				BigDecimal total = new BigDecimal(read);
				Integer temp = 0;
				Map<Integer, Object> hashMap = new HashMap<>();
				for (int i = 0; i < read; i++) {
    
    // 6次
					for (int j = 0; j < 2; j++) {
    
    // 2 颗
						int key = r.nextInt(6) + 1;
						temp = +key;
						if (hashMap.containsKey(key) && j == 1) {
    
    
							hashMap.put(temp, Integer.valueOf(hashMap.get(temp).toString()) + 1);
							temp = 0;
						} else if (j == 1) {
    
    
							hashMap.put(temp, 1);
						}
					}
				}
				if (hashMap != null && hashMap.size() > 0) {
    
    
					for (int i = 2; i < 13; i++) {
    
    
						if (hashMap.get(i) != null) {
    
    
							System.err.println(String.format("数字为%s的概率=%s", i,
									(new BigDecimal(Integer.valueOf(hashMap.get(i).toString()))
											.divide(total, 2, BigDecimal.ROUND_HALF_UP).setScale(2))
													.multiply(new BigDecimal(100))
											+ "%"));
						}
					}
				}
				long e = System.currentTimeMillis();
				System.out.println("耗时：" + (e - s) + "ms");
			} catch (Exception e) {
    
    
				// TODO Auto-generated catch block
				System.err.println("格式有误!");
			}
		}
		scan.close();
	}

}

• Result analysis The
  
  larger the last number of times, the probability of each occurrence is basically the same.