使用含有整数数组的ArrayList-UsingcontainsonanArrayListwithintegerarrays

7  

Arrays can only be compared with Arrays.equals().

数组只能与Arrays.equals()相比较。

You probably want an ArrayList of ArrayLists.

你可能想要一个数组列表。

ArrayList> j = new ArrayList>();
ArrayList w = new ArrayList();
w.add(1); w.add(2);
j.add(w);
ArrayList t = new ArrayList();
t.add(1); t.add(2);
return j.contains(t); // should return true.

7  

The problem here is that arrays don't override Object.equals(Object), So the comparison between two list entries happens with the default equals() implementation

这里的问题是数组不覆盖Object.equals(Object)，因此在默认的equals()实现中，两个列表条目之间的比较发生

// from Object.class
public boolean equals(Object obj) {
return (this == obj);
}

So you have to iterate over the list and check all entries using Arrays.equals(int[], int[]). Here's a Helper method that does this:

因此，必须遍历列表并使用数组检查所有条目。=(int[],int[])。这里有一个帮助方法可以做到这一点:

public static boolean isInList(
    final List list, final int[] candidate){

    for(final int[] item : list){
        if(Arrays.equals(item, candidate)){
            return true;
        }
    }
    return false;
}

Update: Ever since Java 8, this has got a lot simpler:

更新:自从Java 8以来，这变得简单多了:

public static boolean isInList(
        final List list, final int[] candidate) {

    return list.stream().anyMatch(a -> Arrays.equals(a, candidate));
            //  ^-- or you may want to use .parallelStream() here instead
}

0  

You need to iterate through the list and manually check whether an array matches your criteria.

您需要遍历列表并手动检查数组是否匹配您的条件。

public static boolean containsSubArray(List j, int[] sub) {
   for ( int[] arr : j ) {
      if (arr contains elements of sub) {
         return true;
      }
   }
   return false;
}

If you want an exact match, you can make use of Arrays.equals(). I don't think there's a library function to do a contains all on an array though, so you would have to write that yourself if that's what you wanted.

如果你想要精确的匹配，你可以使用Arrays.equals()。我不认为有一个库函数可以在数组中包含所有的元素，所以如果这是你想要的，你必须自己写出来。

0  

"contains" contract checks for equality. So in your case what is failing is equality of int[]. Since Array does not override the equals method from Object you will need a workaround to check for containment.

包含“平等合同检查”。在你的例子中，失败的是int[]的平等。由于数组不覆盖来自对象的equals方法，所以您将需要一个变通方法来检查包含。

If you need to check for containment within the Array then you are left with no choice but to iterate through the ArrayList and do the comparison yourself.

如果您需要检查数组中的容器，那么您将别无选择，只能遍历ArrayList并自己进行比较。

0  

from java api:

从java api:

public boolean contains(Object o)

Returns true if this list contains the specified element. More formally, returns true if and only if this list contains at least one element e such that (o==null ? e==null : o.equals(e)).

如果此列表包含指定的元素，则返回true。更正式地说，如果且仅当此列表包含至少一个元素e (o= null ?)e = = null:o.equals(e))。

since int[] is a primitive, im pretty sure no .equals method exists so its my guess it would always return false.

由于int[]是一个原语，我很确定不存在.equals方法，所以我猜想它总是返回false。

I would recommend a different way of storing the data? maybe with a key of some sort?

我建议用一种不同的方式来存储数据?也许用什么钥匙?

0  

Two java array arrays are equal iff they have the same object reference. Content doesn't matter.

两个java数组是相等的iff，它们具有相同的对象引用。内容不重要。

You're looking for a way to check if they have an equal content. This could help:

你正在寻找一种方法来检查它们是否有相同的内容。这可以帮助:

 Arrays.equals(new int[]{1,2}, new int[]{1,2});   // evaluates to true
 Arrays.equals(new int[]{1,2}, new int[]{2,1});   // evaluates to false (!)

If order shouldn't affect equality, then you will have to implement a static equals method by yourself.

如果顺序不应该影响等式，那么您将不得不自己实现一个静态等号方法。

0  

First they are not the same Object reference, so they are not equal. equals() will return false. For your condition, you will need to implement a method to compare them yourself.

首先它们不是相同的对象引用，所以它们不相等。equals()将返回false。对于您的条件，您将需要实现一个方法来自己比较它们。