硬币翻转代码-尝试计数字符串输出

Question

所以现在我编辑了代码，当有四个头时，计算一行中的头数，当有五个头时计算一行的头数。我有以下正确运行的代码，但它只计算四个数字，而不是五个数字。有人能帮我吗？我将跟踪运行时得到的结果复制并粘贴到我的代码之后。

public class Flip {
    public static void main(String[] args) 
    { 
      final int FLIPS = 100;
     int heads = 0;
     int consecCountfour = 0;
     int consecCountfive = 0;
   System.out.println("Trail Tosses:");
   for (int i = 1; i<= FLIPS; i++)
    {
    if (Math.random() < 0.5) { 
     System.out.print("h"); 
     heads++;
     }
    else {               
       System.out.print("t");
     if (heads == 4) { consecCountfour++; }
     heads = 0;
     if (heads == 5) { consecCountfive++; }
     heads = 0;
     }

  }
  System.out.println("n");
  System.out.print("Count hhhh:"+ consecCountfour);
 System.out.print("     Count hhhhh:" + consecCountfive);
  }
 }

Answer 1

一个不错的解决方案是计算翻转尾部时设置为0的连续头部的数量。

final int FLIPS = 100;
int heads = 0;
int consecCount = 0;
for (int i = 1; i<= FLIPS; i++)
{
    if (Math.random() < 0.5) { 
         System.out.print("h"); 
         heads++;
    }
    else {               
         System.out.print("t");
         if (heads == 4 || heads == 5) { consecCount++; }
         heads = 0;
    }
}
System.out.print(consecCount);