Java中的LinkedList的方法的调用

LinkedList也就是我们在数据结构中链表,这种数据结构具有这样的特性:

  • 分配内存空间不是必须连续的。

  • 插入删除很快,只要修改前后指针就可以了 ,时间复杂度为O(1)。

  • 访问比较慢,必须从第一个元素开始遍历,时间复杂度为O(n)。

在Java中,LinkedList提供了极其丰富的方法,可以模拟链式队列,链式堆栈等数据结构,给我们带来了极大的方便。下面来康康用法。

add(增加一个元素)

boolean add(E e)在链表后添加一个元素,如果成功返回ture,如果不成功,则返回false。

void addFirst(E e)在链表头部添加一个元素。

void addLast(E e)在链表尾部添加一个元素。

void add(int index ,E element)在指定位置添加一个元素。

代码演示:

public class LinkedListMethodsDemo {
    public static void main(String[] args) {
        LinkedList<String> linkedList = new LinkedList<>();

        linkedList.add("first");
        linkedList.add("second");
        linkedList.add("third");
        System.out.println(linkedList);

        linkedList.addFirst("addFirst");
        System.out.println(linkedList);

        linkedList.addLast("addLast");
        System.out.println(linkedList);

        linkedList.add(2, "addByIndex");
        System.out.println(linkedList);
    }
}

结果:

[first, second, third]
[addFirst, first, second, third]
[addFirst, first, second, third, addLast]
[addFirst, first, addByIndex, second, third, addLast]

remove(移出):

E remove() 删除链表中第一个元素。

boolean remove(Object o)移除链表中指定的元素。

E remove(int index):移除链表中指定位置的元素。

E removeFirst() 移除链表中的第一个元素,与remove类似。

E removeLast()移出链表中的最后一个元素。

boolean removeFirstOccurrence(Object o)移除链表中第一次元素出现所在的位置。

boolean removeLastOccurrence(Object o)移除链表中最后一次元素出现所在的位置。

代码演示

public class LinkedListMethodsDemo {
    public static void main(String[] args) {
        LinkedList<String> linkedList = new LinkedList<>();

        linkedList.add("first");
        linkedList.add("second");
        linkedList.add("second");
        linkedList.add("third");
        linkedList.add("four");
        linkedList.add("five");
        System.out.println(linkedList);

        linkedList.remove();
        System.out.println("remove: " + linkedList);

        linkedList.remove("second");
        System.out.println("remove(Object): " + linkedList);

        linkedList.remove("six");
        System.out.println("remove(Object) not exist: " + linkedList);

        linkedList.remove(2);
        System.out.println("remove(index): " + linkedList);

        linkedList.removeFirst();
        System.out.println("removeFirst: " + linkedList);

        linkedList.removeLast();
        System.out.println("removeLast:" + linkedList);

        System.out.println("----------------------------------");
        linkedList.clear();

        linkedList.add("first");
        linkedList.add("second");
        linkedList.add("first");
        linkedList.add("third");
        linkedList.add("first");
        linkedList.add("five");
        System.out.println(linkedList);

        linkedList.removeFirstOccurrence("first");
        System.out.println("removeFirstOccurrence: " + linkedList);

        linkedList.removeLastOccurrence("first");
        System.out.println("removeLastOccurrence: " + linkedList);
    }
}

结果:

[first, second, second, third, four, five]
remove: [second, second, third, four, five]
remove(Object): [second, third, four, five]
remove(Object) not exist: [second, third, four, five]
remove(index): [second, third, five]
removeFirst: [third, five]
removeLast:[third]
----------------------------------
[first, second, first, third, first, five]
removeFirstOccurrence: [second, first, third, first, five]
removeLastOccurrence: [second, first, third, five]

get(获取)

E get(int index) 按照下边获取元素

E getFirst():获取第一个元素

E getLast():获取最后一个元素。

代码演示

public class LinkedListMethodsDemo {
    public static void main(String[] args) {
        LinkedList<String> linkedList = new LinkedList<>();

        linkedList.add("first");
        linkedList.add("second");
        linkedList.add("second");
        linkedList.add("third");
        linkedList.add("four");
        linkedList.add("five");
        System.out.println(linkedList);

        linkedList.get(3);
        System.out.println("get(index): " + linkedList.get(3));

        linkedList.getFirst();
        System.out.println("getFirst: " + linkedList.getFirst());

        linkedList.getLast();
        System.out.println("getLast: " + linkedList.getLast());

        System.out.println(linkedList);
    }
}

结果:

[first, second, second, third, four, five]
get(index): third
getFirst: first
getLast: five
[first, second, second, third, four, five]

push,pop,poll

void push(E e) ,与addFirst一样,实际上他就是addFirst,在链表头部添加一个元素。

E pop():与removeFirst一样,实际上就是removeFirst。移除链表中的第一个元素.

E poll():查询并移除第一个元素;

代码演示:

public class LinkedListMethodsDemo {
    public static void main(String[] args) {
        LinkedList<String> linkedList = new LinkedList<>();

        linkedList.push("first");
        linkedList.push("second");
        linkedList.push("second");
        linkedList.push("third");
        linkedList.push("four");
        linkedList.push("five");
        System.out.println("linkedList: " + linkedList);

        System.out.println("pop: " + linkedList.pop());
        System.out.println("after pop: " + linkedList);

        System.out.println("poll: " + linkedList.poll());
        System.out.println("after poll: " + linkedList);
    }
}

结果:

linkedList: [five, four, third, second, second, first]
pop: five
after pop: [four, third, second, second, first]
poll: four
after poll: [third, second, second, first]

通过代码示例可以看出:push,pop的操作已经很接近stack(栈)的操作了。

如果链表为空的时候,看看poll与pop是啥区别:

public class LinkedListMethodsDemo {
    public static void main(String[] args) {
        LinkedList<String> linkedList = new LinkedList<>();

        System.out.println("poll: " + linkedList.poll());
        System.out.println("pop: " + linkedList.pop());
    }
}

结果:

poll: null
Exception in thread "main" java.util.NoSuchElementException
    at java.util.LinkedList.removeFirst(LinkedList.java:270)
    at java.util.LinkedList.pop(LinkedList.java:801)
    at org.ks.algorithm.LinkedListMethodsDemo.main(LinkedListMethodsDemo.java:13)
    at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
    at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:62)
    at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
    at java.lang.reflect.Method.invoke(Method.java:483)
    at com.intellij.rt.execution.application.AppMain.main(AppMain.java:144)

可以看出poll返回null,而pop则产生异常。

peek

E peek() : 获取第一个元素但是不移除。

E peekFirst : 获取第一个元素但是不移除。

E peekLast : 获取最后一个元素但是不移除。

代码演示:

public class LinkedListMethodsDemo {
    public static void main(String[] args) {
        LinkedList<String> linkedList = new LinkedList<>();

        linkedList.push("first");
        linkedList.push("second");
        linkedList.push("second");
        linkedList.push("third");
        linkedList.push("four");
        linkedList.push("five");
        System.out.println("linkedList: " + linkedList);

        System.out.println("peek: " + linkedList.peek());
        System.out.println("peekFirst: " + linkedList.peekFirst());
        System.out.println("peekLast: " + linkedList.peekLast());

        System.out.println("linkedList: " + linkedList);
    }
}

输出:

linkedList: [five, four, third, second, second, first]
peek: five
peekFirst: five
peekLast: first
linkedList: [five, four, third, second, second, first]

如果没找到对应的元素,统统输出null:

public class LinkedListMethodsDemo {
    public static void main(String[] args) {
        LinkedList<String> linkedList = new LinkedList<>();

        System.out.println("linkedList: " + linkedList);
        System.out.println("peek: " + linkedList.peek());
        System.out.println("peekFirst: " + linkedList.peekFirst());
        System.out.println("peekLast: " + linkedList.peekLast());
    }
}

输出:

linkedList: []
peek: null
peekFirst: null
peekLast: null

offer

boolean offer(E e) : 在链表尾部插入一个元素。

boolean offerFirst(E e) : 与addFirst一样,实际上它就是addFirst;

boolean offerLast(E e) : 与addLast一样,实际上它就是addLast;

代码演示:

public class LinkedListMethodsDemo {
    public static void main(String[] args) {
        LinkedList<String> linkedList = new LinkedList<>();

        linkedList.push("first");
        linkedList.push("second");
        linkedList.push("second");
        linkedList.push("third");
        linkedList.push("four");
        linkedList.push("five");
        System.out.println("linkedList: " + linkedList);

        linkedList.offer("six");
        System.out.println("linkedList: " + linkedList);

        linkedList.offerFirst("zero");
        System.out.println("linkedList: " + linkedList);

        linkedList.offerLast("seven");
        System.out.println("linkedList: " + linkedList);
    }
}

输出:

linkedList: [five, four, third, second, second, first]
linkedList: [five, four, third, second, second, first, six]
linkedList: [zero, five, four, third, second, second, first, six]
linkedList: [zero, five, four, third, second, second, first, six, seven]

LinkedList中常用的方法基本都列出来了,当然还有一些其他的例子,这里就一起演示了:

boolean contains(Object o) : 如果此列表包含指定元素,则返回 true。

E element() :获取但不移除此列表的头(第一个元素)。

E set(int index ,E element) : 将此列表中指定位置的元素替换为指定的元素。

List<E> subList(int fromIndex,int toIndex) :返回列表中指定的 fromIndex(包括 )和 toIndex(不包括)之间的部分视图。(如果 fromIndex 和 toIndex 相等,则返回的列表为空)。

public class LinkedListMethodsDemo {
    public static void main(String[] args) {
        LinkedList<String> linkedList = new LinkedList<>();

        linkedList.push("first");
        linkedList.push("second");
        linkedList.push("second");
        linkedList.push("third");
        linkedList.push("four");
        linkedList.push("five");
        System.out.println("linkedList: " + linkedList);

        System.out.println("linkedList.contains(\"second\"): " + linkedList.contains("second"));  
        System.out.println("linkedList.contains(\"six\"): " + linkedList.contains("six"));
        System.out.println("linkedList.element(): " + linkedList.element());
        System.out.println("linkedList: " + linkedList);
        System.out.println("linkedList.set(3, \"set\"): " + linkedList.set(3, "set"));
        System.out.println("linkedList: " + linkedList);
        System.out.println("linkedList.subList(2,4): " + linkedList.subList(2,4));
        System.out.println("linkedList: " + linkedList);
    }
}

输出:

linkedList: [five, four, third, second, second, first]
linkedList.contains("second"): true
linkedList.contains("six"): false
linkedList.element(): five
linkedList: [five, four, third, second, second, first]
linkedList.set(3, "set"): second
linkedList: [five, four, third, set, second, first]
linkedList.subList(2,4): [third, set]
linkedList: [five, four, third, set, second, first]
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