錯誤：嘗試返回ArrayList時不兼容的類型

Question

我正在嘗試解決Project Euler上的Problem 12。 我有一個如何完成這個問題的想法，但是我遇到了一個 錯誤。我已經搜索瞭如何從不同的問題中使用ArrayList，但是我仍然面臨問題 。

import java.util.ArrayList; 

public class Level_12 { 
/* 
The sequence of triangle numbers is generated by adding the natural numbers. 
So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. 

The first ten terms would be: 

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ... 

Let us list the factors of the first seven triangle numbers: 

1: 1 
3: 1,3 
6: 1,2,3,6 
10: 1,2,5,10 
15: 1,3,5,15 
21: 1,3,7,21 
28: 1,2,4,7,14,28 
We can see that 28 is the first triangle number to have over five divisors. 

What is the value of the first triangle number to have over five hundred divisors? 
*/ 

public static ArrayList<Long> check(long num) { 
    ArrayList<Long> divisors = new ArrayList<Long>(); 
    for (long o = 1; o <= Math.sqrt(num); o++) 
    if (num % o == 0) { 
     divisors.add(o); 
     System.out.println(o + " is a current divisor of " + num); 
    } 
    for (Long m : divisors) { 
    return m; 
    } 
} 

public static void main(String[] args) { 
    long triangle = 0; //Triangle number 
    long total = 500; //Total divisors 
    long currenttotal = 0; //Amount of divisors 
    long i = 0; //Just to itterate 

    while (currenttotal <= total) { 
     if (check(i) > currenttotal) { //Finding if the 
      triangle = i; 
      if (currenttotal == total) break; 
     } 
     i++; 
    } 
    System.out.println("The value of the first triangle number to have over 500 divisors is " + triangle); 

    } 
} 

編輯：

與最終代碼固定爲

import java.util.ArrayList; 

public class Level_12 { 
/* 
The sequence of triangle numbers is generated by adding the natural numbers. 
So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. 

The first ten terms would be: 

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ... 

Let us list the factors of the first seven triangle numbers: 

1: 1 
3: 1,3 
6: 1,2,3,6 
10: 1,2,5,10 
15: 1,3,5,15 
21: 1,3,7,21 
28: 1,2,4,7,14,28 
We can see that 28 is the first triangle number to have over five divisors. 

What is the value of the first triangle number to have over five hundred divisors? 
*/ 

public static ArrayList<Long> check(long num) { 
    ArrayList<Long> divisors = new ArrayList<Long>(); 
    for (long o = 1; o <= Math.sqrt(num); o++) 
    if (num % o == 0) { 
     divisors.add(o); 
     System.out.println(o + " is a current divisor of " + num); 
    } 
    return divisors; 
    } 


public static void main(String[] args) { 
    long triangle = 0; //Triangle number 
    long total = 500; //Total divisors 
    long currenttotal = 0; //Amount of divisors 
    long i = 0; //Just to itterate 

    while (currenttotal <= total) { 
     if (check(i).size() > currenttotal) { //Finding if the amount of divisors is larger than current divisors 
      triangle = i; 
      if (currenttotal == total) break; 
     } 
     i++; 
} 
    System.out.println("The value of the first triangle number to have over 500 divisors is " + triangle); 

    } 
} 

Answer 1

罪魁禍首是這樣的：

for (Long m : divisors) { 
    return m; 
    } 

在此，而該方法簽名指定返回的ArrayList

public static **ArrayList<Long>** check(long num) 

你，你是返回一個長整型值需要更改方法簽名或根據您想要完成的內容返回適當的值。

Answer 2

你在你的check()方法

Answer 3

返回的Long，而不是ArrayList<Long>返回Long，而返回類型方法是ArrayList<Long>。因此，更新回報鍵入如下：

public static Long check(long num) { 
    ArrayList<Long> divisors = new ArrayList<Long>(); 
    for (long o = 1; o <= Math.sqrt(num); o++) 
    if (num % o == 0) { 
     divisors.add(o); 
     System.out.println(o + " is a current divisor of " + num); 
    } 
    for (Long m : divisors) { 
    return m; 
    } 
} 

Answer 4

您只能將m的一半因子放入列表中。如果o除以m，那麼m/o也除m。

/** 
* Return a list of the divisors of an integer. If it is a perfect square, the square 
* root will be repeated. The list will not be in order. 
* @param num The number whose divisors to find. 
* @return A list of the number's positive divisors. 
*/ 
public List<Long> divisors(long num) { 
    final List<Long> divisors = new ArrayList<>(); 
    final long limit = Math.sqrt(num); 
    for (long o = 1L; o <= limit; o++) 
    if (num % o == 0) { 
     divisors.add(o); 
     divisors.add(num/o); 
    } 
    return divisors; 
} 

/** 
* Return the nth triangular number. 
* @param n The index of the desired triangular number. 
* @return The n-th triangular number 1 + 2 + ... + n 
*/ 
public long triangle(long n) { 
    return n * (n + 1)/2; 
}