Java：鏈表 - (boolean [head=true, tail false])

Question

條件一：如果你輸入了多於或等於10個直頭並且能夠在輸入中顯示，那么返回是“Streak is found”

條件 2：如果您輸入的直頭少於 10 個，則返回“Streak is broken”

但是，我遇到了條件 1 的問題，它沒有執行到輸出。

編碼：

import java.util.LinkedList;
import java.util.Iterator;
import java.util.Scanner;

public class LinkedListProgram2
{    
public static void main (String [] args)
{
    Scanner input = new Scanner (System.in);
    LinkedList<String> cointoss = new LinkedList<String>();
    boolean head = true;
    boolean tail = false;
    boolean streak = true;
    int streakcount = 0;

    System.out.println ("Welcome to the Program #2 ");
    //ask for the boolean value. It can be head and tail or true and false.
    System.out.print ("\nEnter the boolean value (head=true, tail=false): ");
    
    for (int i = 0; i<18; i++)
    {
        cointoss.add(input.next());        
    }
    
    Iterator<String> it = cointoss.iterator();
    while (it.hasNext())
    {
        if(streakcount >= 10)
        {
            streak = true;
            System.out.println ("Streak is found! ");
            break;
        }
        else if(streakcount < 10)
        {
            streak = false;
            System.out.println ("Streak is broken! ");
            break;
        }
    }
}
}

Answer 1

至少需要添加2個邏輯。

1. 找到 true 時增加連續計數

2. 發現假時重置計數器

您可以在while循環內使用外部 if 條件為每個實例打印broken ，或者讓它在循環結束時打印在外面

    while (it.hasNext()) {
        String val = it.next();
        if (val.equals("true"))
            streakcount++;
        else
            streakcount = 0;
        
        if (streakcount >= 10) {
            streak = true;
            System.out.println("Streak is found! ");
            break;
        }
    }
    if (!streak) {
        System.out.println("Streak is broken! ");
    }

Answer 2

您錯過了代碼中的某些內容，您需要檢查輸入值，並增加streakcount ，未經測試的示例代碼：

while (it.hasNext()) {
    String val = it.next();
    if (val.equals("true")) {
        streakcount++;
        if (streakcount >= 10) {
            streak = true;
            System.out.println ("Streak is found! ");
            break;
        }
    }
    else if (val.equals("false")) {
        streak = false;
        System.out.println ("Streak is broken! ");
        break;
    }
}

還有更多操作要做，檢查不同的輸入值，或者如果不是從起始數組開始，是否需要查找條紋...

Answer 3

根據您的描述，這實際上是一個計算特定字符串的小程序，甚至不需要 LinkedList 我修改了您的代碼，現在它應該滿足您提出的兩個條件

    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        LinkedList<String> cointoss = new LinkedList<String>();
        boolean head = true;
        boolean tail = false;
        boolean streak = true;
        int streakcount = 0;

        System.out.println("Welcome to the Program #2 ");
        // ask for the boolean value. It can be head and tail or true and false.
        System.out.println("Enter the boolean value (head=true, tail=false): ");

        for (int i = 0; i < 18; i++) {
            String next = input.next();
            if (next.equals("true") || next.equals("head")) {
                streakcount++;
            }
            cointoss.add(next);
        }

        if (streakcount >= 10) {
            System.out.println("Streak is found! ");
        } else {
            System.out.println("Streak is broken! ");
        }
    }