硬幣翻轉代碼 - 試圖數出字符串輸出

Question

所以，現在我編輯的代碼，當有四個時，計算行中的頭數和五個連續時的頭數。我有以下代碼可以正確運行，但它只計算四個數字而不是五個數字。誰能幫我？我複製並粘貼了我的代碼之後運行的結果。

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

一個不錯的解決方案是計算連續的磁頭數量，當尾部被翻轉時，連續的磁頭數量被設置回0。

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);