我有这个代码打印出一些骰子的直方图。我的问题是,我想打印第二个旁边的第一个直方图,目前我的输出看起来像这个,但我想把星星放在右侧的滚动次数如下:2:3次/// 3:1次/依此类推。
public static void main(String[] args) {
// TODO code application logic here
System.out.print("Please enter how many times to roll the dice: ");
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int [] rolls = new int[n];
Random r1 = new Random();
Random r2 = new Random();
int dice1;
int dice2;
int [] t = new int [13];
for (int roll=0; roll < rolls.length; roll++)
{
dice1 = r1.nextInt(6)+1;
dice2 = r2.nextInt(6)+1;
System.out.println(roll + " : I rolled a " + dice1 + " and a " + dice2);
int sum;
sum = dice1 + dice2;
if (sum == 2)
t[0]++;
if (sum == 3)
t[1]++;
if (sum == 4)
t[2]++;
if (sum == 5)
t[3]++;
if (sum == 6)
t[4]++;
if (sum == 7)
t[5]++;
if (sum == 8)
t[6]++;
if (sum == 9)
t[7]++;
if (sum == 10)
t[8]++;
if (sum == 11)
t[9]++;
if (sum == 12)
t[10]++;
}
System.out.println("Histogram of rolls:" );
String star ="*";
int [] h= {t[0], t[1],t[2], t[3],t[4], t[5], t[6], t[7],t[8], t[9],t[10]};
for (int i=0; i <h.length; i++)
{
for(int j = 0; j < h[i]; j++)
System.out.print( star);
System.out.println();
}
System.out.println("Histogram of rolls:" );
System.out.println( "2 : " + t[0] + " times");
System.out.println("3 : " + t[1] + " times");
System.out.println("4 : " + t[2] + " times");
System.out.println("5 : " + t[3] + " times");
System.out.println("6 : " + t[4] + " times");
System.out.println("7 : " + t[5] + " times");
System.out.println("8 : " + t[6] + " times");
System.out.println("9 : " + t[7] + " times");
System.out.println("10 : " + t[8] + " times");
System.out.println("11 : " + t[9] + " times");
System.out.println("12 : " + t[10] + " times");
}
}
答案 0 :(得分:1)
将代码的最后一部分更改为:
...
System.out.println("Histogram of rolls:" );
String star ="*";
// No point in using "int [] h"
for (int i=0; i < t.length; i++) {
// Placing the logic for printing the text inside the loop
// is how you use arrays
System.out.print( (i+2) + " : " + t[0] + " times");
for(int j = 0; j < t[i]; j++) {
System.out.print(star);
}
System.out.println();
}
}
更重要的是,您可能缺少使用数组的关键优势:只要有可能,就应该使用循环,而不是硬编码顺序逻辑。例如,if语句的大集群:
if (sum == 2)
t[0]++;
if (sum == 3)
t[1]++;
if (sum == 4)
t[2]++;
if (sum == 5)
t[3]++;
if (sum == 6)
t[4]++;
if (sum == 7)
t[5]++;
if (sum == 8)
t[6]++;
if (sum == 9)
t[7]++;
if (sum == 10)
t[8]++;
if (sum == 11)
t[9]++;
if (sum == 12)
t[10]++;
可以简单地简化为:
t[sum-2]++;
答案 1 :(得分:1)
建议:
1)你根本不需要数组h []
2)您需要在同一个循环内打印直方图(“***”)的每一行和“直方图卷数”的每一行
3)我会创建一个变量“String asterisks = "************";
并使用subString。()副本打印正确数量的星号,而不是”for(...)“循环。
PS:
String.subString()是我在other question
中尝试建议的内容