java 如何最有效地在 LinkedList 中查找最大和最小元素？
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How to find max and min element in a LinkedList most efficiently?
提问by h4ck3d
I have a linked list in Java ,say , LinkedList<T> list = new LinkedList<T>();and I need to find the max / min element in it most efficiently , how can I do it?
比如说，我有一个 Java 链表，LinkedList<T> list = new LinkedList<T>();我需要最有效地在其中找到最大/最小元素，我该怎么做？
How can I use Collections.max()function to find the max element from my linked list? What is the time complexity of this function ?
如何使用Collections.max()函数从链表中查找最大元素？这个函数的时间复杂度是多少？
回答by Peter Lawrey
The time complexity is O(n)if you want it to be lower e.g. O(1)you need to use a different data structure such as a TreeSet.
时间复杂度是O(n)如果您希望它更低，例如O(1)您需要使用不同的数据结构，例如 TreeSet。
how can i use Collections.max() for LinkedList
我如何将 Collections.max() 用于 LinkedList
List<Integer> list = ...
Integer max = Collections.max(list);
回答by Adam Sznajder
ArrayList<Integer> numbers = new ArrayList<Integer>();
/* fill with values */
Integer max = Collections.max(numbers);
回答by Amit Deshpande
Java Doc
Java文档
This method iterates over the entire collection, hence it requires time proportional to the size of the collection. (O(n))
此方法迭代整个集合，因此它需要与集合大小成正比的时间。（在））
Usage
用法
public class Test
{
public static void main(String args[])
{
// create an LinkedList object
LinkedList<Integer> list = new LinkedList<Integer>();
// Add elements to LinkedList
list.add(1);
list.add(2);
list.add(3);
list.add(4);
list.add(5);
//Use Max Method
System.out.println(Collections.max(list));
}
}
回答by Ritam Chakraborty
You can also use stream.max()to get the max value of a LinkedList
您还可以使用stream.max()获取 LinkedList 的最大值
LinkedList<Integer> linkedList = new LinkedList<>();
linkedList.add(1); linkedList.add(2);
linkedList.add(3); linkedList.add(4);
linkedList.stream().max(Integer::compareTo).ifPresent(System.out::println);
Output: 4
