java coinFlip磁头条纹输出

Question

我应该把一枚硬币掷100次，找出最长的头像，然后输出，但到目前为止，它只给出了总共有多少头像。我试过很多方法，但都找不到解决办法

public class LongestStreak extends ConsoleProgram
{
    public static final int FLIPS = 100;

    public void run()
    {
        int h = 0;
        int t = 0;
        boolean wasHeads = false;
        boolean isHeads = false;
        int streak = 0;
        int ih = 0;

        for (int i = 0; i < FLIPS; i++) {
            int coinFlip = Randomizer.nextInt(1, 2);

            if (coinFlip == 2) {
                System.out.println("Heads");
                h++;
                ih++;
                isHeads = true;

                if (ih > 1) {
                    wasHeads = true;
                }
                if ((isHeads = true) && (wasHeads = true)) {   
                    streak++;
                } else {
                    streak = 0;
                }
            } else if (coinFlip == 1) {
                System.out.println("Tails");
                t++;  
                isHeads = false;
                ih = 0;
            }
        }

        System.out.println(streak);
    }
}

Answer 1

您需要大大简化代码

public class LongestStreak extends ConsoleProgram
{
    public static final int FLIPS = 100;

    public void run()
    {
        int wasHeads = 0;
        int streak = 0;
        int bestStreak = 0;

        for (int i = 0; i < FLIPS; i++) {
            int coinFlip = Randomizer.nextInt(1, 2);

            if (coinFlip == 2) {
                streak++;
                wasHeads = 1;
            } else if (coinFlip == 1) {
                wasHeads = 0;
                if (steak > bestStreak) {
                    bestStreak = streak;
                }
                streak = 0;
            }
        }
        System.out.println(streak);
    }
}

结束时重置条纹（coinFlip为1），但首先检查它是否优于之前的最佳条纹，如果是，则相应地更新最佳条纹。否则，只需重复streak并将每个头部的洗头数保持为1。注意这里不需要布尔值；1或0整数在字面上和逻辑上是相同的

Answer 2

感谢您的帮助，修复了我的代码：）

public class LongestStreak extends ConsoleProgram

{ 公共静态最终整数翻转=100

public void run()
{
    int h = 0;
    int t = 0;
    boolean wasHeads = false;
    boolean isHeads = false;
    int longest = 0;
    int streak = 0;
    int ih = 0;

    for (int i = 0; i < FLIPS; i++) {

        int coinFlip = Randomizer.nextInt(1, 2);

        if (coinFlip == 2) { //heads

            System.out.println("Heads");

            if (wasHeads = true) {
                streak++;
            } else {
                streak = 0;
            }

            wasHeads = true;
        } else if(coinFlip == 1) { //tails

            System.out.println("Tails");

            t++;

            isHeads = false;
            if (streak > longest){
                longest = streak;
            }

            streak = 0;

            ih = 0;

        }

    }

    System.out.println("Longest streak of heads: " + longest);
}

}

Answer 3

首先也是最重要的一点是，在这里修改您的行以便更容易阅读：}} else if (coinFlip == 1) {
（这两个紧括号之间需要换行符）

我会考虑设置一些更清楚的变量名。例如：totalHeads，totalTails。我不太明白你的ih变量应该是什么，但是我用来解决这个问题的变量名是thisStreak和bestStreak。is/washeads应该是不必要的。我认为这应该足以推动你找到解决方案

Answer 4

您可以稍微简化代码，wasHeads和isHeads实际上并不需要。此外，更具描述性的变量名使代码更易于理解。比如说：

public void run() {
    int highestNumberOfConsecutiveHeads = 0;
    int currentNumberOfConsecutiveHeads = 0;
    for (int i = 0; i < 100; i++) {
        if (Randomizer.nextInt(1, 2) == 2) {
            currentNumberOfConsecutiveHeads++;
            if (currentNumberOfConsecutiveHeads > largestNumberOfConsecutiveHeads) {
                highestNumberOfConsecutiveHeads = currentNumberOfConsecutiveHeads;
            }
        } else {
            currentNumberOfConsecutiveHeads = 0;
        }
    }
    System.out.println("Longest streak: " + highestnumberOfConsecutiveHeads);
}