blocks|key|4501711|text|通过丢弃旧的数组，分配一个全新的数组，然后将旧数组中的值复制到新的数组中，ArrayList可以增长到受可用内存限制的任何大小。此操作可以对任何给定的插入(即amortized+cost+is+O(1)+)使用O(n)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4501712|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/q/2182597/335858^0|11|9|2P|4|2X|4|27|M|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]|$9|Q|A|R|B|C]|$9|S|A|T|B|C]]|D|@$9|U|A|V|1|W]]|E|$]]|$1|F|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|G|$H|$5|I|J|K|E|$L|M]]]]

<code>ArrayList</code> can grow to any size limited by the available memory by discarding its old array, allocating a brand-new one, and copying the values from the old array into the new one. This operation can take <code>O(n)</code> for any given insertion, the <a href="https://stackoverflow.com/q/2182597/335858">amortized cost is <code>O(1)</code></a>.

blocks|key|2864735|text|所以我假设ArrayLists存储在堆中的连续位置|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2864736|2864737|通常是这样的，但没有什么能阻止JVM做一些不同的事情(+JVM规范中没有规定数组必须存储在连续的块中)。|2864738|另请参见this+related+post。|offset|length|2864739|2864740|如果堆中存储了阻止ArrayList增长的另一个对象，该怎么办？|blockquote|2864741|2864742|2864743|arraylist对象将请求JVM给它更多的空间，JVM将根据自己的意愿分配空间(如果不能，则抛出一个OutOfMemoryException+)。|2864744|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/10224888/java-are-1-d-arrays-always-contiguous-in-memory^0|0|0|0|4|H|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]|$1|C|3|D|5|6|7|10|8|@]|9|@]|A|$]]|$1|E|3|F|5|6|7|11|8|@]|9|@$G|12|H|13|1|14]]|A|$]]|$1|I|3|-4|5|6|7|15|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|16|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|17|8|@]|9|@]|A|$]]|$1|N|3|-4|5|6|7|18|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|19|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|1A|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X]]]]

<blockquote>
 So I assume ArrayLists are stored in the heap in contiguous locations
</blockquote>

It is generally the case but nothing prevents a JVM from doing something different (nothing in the JVM specifications says that an array has to be stored in continuous blocks).

See also <a href="https://stackoverflow.com/questions/10224888/java-are-1-d-arrays-always-contiguous-in-memory">this related post</a>.

<blockquote>
 what if it happens that there is another object that was stored in the heap that would prevent the ArrayList from growing?
</blockquote>

The arraylist object will ask the JVM to give it more space, which the JVM will allocate as it wishes (or throw an OutOfMemoryException if it can't).

blocks|key|4501736|text|查看源代码(感谢@GilbertoGaxiola)，最初的ArrayList创建了一个大小为10的object[]，因为它需要更多的空间，所以它将旧数组的大小加倍，并将旧数组的内容复制到新数组中。在这个数组变得巨大之前，它不需要填满很多次，这使得它成为一个很好的实现。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4501737|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Looking at the source code (thanks @GilbertoGaxiola) the initial ArrayList creates an object[] of size 10. As it needs more space, it doubles the size and copies the contents of the old array into the new array. It doesn't take very many times filling up before this array is huge, which makes this a great implementation.

blocks|key|4592334|text|get，所以我假设ArrayLists存储在堆中的连续位置(这就是他们使用O(1)+|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4592335|方法的原因)。|blockquote|4592336|4592337|4592338|所有其他答案在这里都是正确的，但我想解决您在这里做出的一个不正确的假设：|4592339|ArrayList.get()为O(1)不是因为数组在内存中的存储方式，而是因为元素访问的成本是恒定的，并且与ArrayList中的元素数量不成比例。|offset|length|style|CODE|4592340|O(1)这里并不意味着查找可以在一次操作中完成-它意味着复杂度相对于后备数组的大小是恒定的。|4592341|当您调用ArrayList.get(i)时，JVM仍然需要将数组索引转换为虚拟内存地址，这涉及到您的操作系统将地址转换为实际地址-但这是O(1)，因为如果数组有1个元素或5000个元素，复杂度是相同的。|4592342|相比之下，LinkedList.get(i)不是O(1)，因为LinkedList类需要遍历每个节点，直到找到节点i+-复杂性与列表的整体大小和i的大小成比例。|4592343|数组存储在连续位置的假设也是不正确的。|4592344|entityMap^0|0|0|0|0|0|0|F|0|0|4|G|0|5|H|1L|1|20|1|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$]]|$1|E|3|-4|5|6|7|10|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|11|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|12|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|13|8|@$K|14|L|15|M|N]]|9|@]|A|$]]|$1|O|3|P|5|6|7|16|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|17|8|@$K|18|L|19|M|N]]|9|@]|A|$]]|$1|S|3|T|5|6|7|1A|8|@$K|1B|L|1C|M|N]|$K|1D|L|1E|M|N]|$K|1F|L|1G|M|N]]|9|@]|A|$]]|$1|U|3|V|5|6|7|1H|8|@]|9|@]|A|$]]|$1|W|3|-4|5|6|7|1I|8|@]|9|@]|A|$]]]|X|$]]

<blockquote>
 So I assume ArrayLists are stored in the heap in contiguous locations (Thats the reason they have an O(1) get method).
</blockquote>

All of the other answers are here on correct, but I want to address an incorrect assumption you've made here:

<code>ArrayList.get()</code> is O(1) not because of how the array is stored in memory, but because the cost of the element access is constant and not proportional to the number of elements in the ArrayList.

O(1) here does not mean that a lookup can be done in a single operation - it means that the complexity is constant with respect to the size of the backing array.

When you call <code>ArrayList.get(i)</code> the JVM still needs to translate the array index into a virtual memory address, which involves your OS translating the address into a real one - but it's O(1) because the complexity is the same if the array has 1 element or 5000.

In comparison, <code>LinkedList.get(i)</code> is not O(1) because the LinkedList class needs to walk through each node until it finds node <code>i</code> - the complexity is in proportion to the overall size of the list and the magnitude of <code>i</code>.

The assumption that the array is stored in contiguous locations is also incorrect.

blocks|key|4501881|text|最后一个解释是完全正确的。os最终必须将索引转换为真实地址。因此，对于所有索引，转换所用的时间都是恒定的。关于存储，第二个解释是正确的，如果你必须存储更多的元素，那么它的大小会增加一倍。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4501882|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

The last explanation is absolutely correct. The os has to ultimately translate the index to a real address. So the time taken for translation is constant for all indexes. Regarding storage the secnd last explanation is correct, the size just doubles if you have to store more elements .

I believe there is a difference between LinkedLists and ArrayLists. ArrayLists are nothing but dynamic arrays. So I assume ArrayLists are stored in the heap in contiguous locations (Thats the reason they have an O(1) get method). The question is, what if it happens that there is another object that was stored in the heap that would prevent the ArrayList from growing? How is it implemented in this case? If the remaining of the ArrayList is stored in other non-conrigous area of the heap, the get method wont be an O(1).

For instance, assume that there is an object in memory location 10. After that, an ArrayList is created at memory location 5. For simplicity assume each element in the array list is just one byte. This means the ArrayList can grow to 5 elements only.

How is an ArrayList data structure indexed and dynamic at the same time? How is it implemented?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我相信LinkedLists和ArrayLists是有区别的。ArrayLists只不过是动态数组。因此，我假设ArrayLists存储在堆中的连续位置(这就是它们具有O(1) get方法的原因)。问题是，如果堆中存储了另一个阻止ArrayList增长的对象，该怎么办？在这种情况下它是如何实现的？如果ArrayList的其余部分存储在堆的其他非冲突区域中，则get方法将不是O(1)。例如，假设在内

问如何同时索引ArrayList数据结构和动态数据结构？它是如何实现的？
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