删除链接列表中所有出现的元素

Question

在我理解数据结构的过程中，我开始在java中实现它们。deleteAll的时间复杂度为O(n + n^2)。如何改进deleteAll方法？

/*
 * Singly linked list
 */
package linkedlisttest;

class Node {

    int data;
    Node next;

    public Node(int data) {
        this.data = data;
    }

}

class LinkedList {

    Node head;
    int size;

    /**
     *
     * @param data element to add to list 
     * Time Complexity : O(n)
     */
    public void add(int data) {
        if (head == null) {
            head = new Node(data);
            size += 1;
            return;
        }

        Node current = head;
        while (current.next != null) {
            current = current.next;
        }
        current.next = new Node(data);
        size += 1;
    }

    /**
     *
     * @return size of list 
     * Time Complexity: O(1) 
     * This is because we use a class
     * variable size to keep track of size of linked list
     */
    public int getSize() {
        return size;
    }
    /**
     * 
     * @param data element to insert 
     * @param index position at which to insert the element (zero based)
     * Time Complexity : O(n)
     */
    public void add(int data, int index) {

        if (index > getSize()) {
            return; // invalid position
        }

        Node current = head; //iterate through whole list 
        int pos = 0;
        Node newNode = new Node(data);

        if (index == 0) // special case, since its a single reference change!
        {
            newNode.next = head;
            head = newNode; // this node is now the head
            size += 1;
            return;
        }
        while (current.next != null) {
            if (pos == index - 1) {
                break;
            }
            pos++;
            current = current.next;
        }
        // These are 2 reference changes, as compared to adding at index 0
        newNode.next = current.next; // here we are changing a refernce
        current.next = newNode; // changing a reference here as well
        size += 1;

    }
    /**
     * Find the first occurrence of an element  
     * @param data element to find
     * @return index at which element is found , -1 if not exist
     * Time Complexity: O(n)
     */
    public int find(int data) {
        Node current = head;
        int pos = 0;
        int index = -1;
        if(head == null) { //empty list
            return index;
        }
        while(current != null) {
            if (current.data == data) {
                index = pos;
                break;
            }
            pos++;
            current = current.next;
        }
        return index;
    }

    /**
     * Delete the first occurrence of data 
     * @param data element to delete 
     * Time complexity : O(n)
     */
    public void delete(int data) {
        Node current = head;
        if (head == null) { // list is empty
            return;
        }
        if(head.data == data) { // if we want to delete the head , make next node head
            head = head.next;
            size -= 1;
            return;
        }
        while(current.next != null) {
            if (current.next.data == data) {
                current.next = current.next.next;
                size -= 1;
                return;
            }
            current = current.next;
        }
    }

    /**
     * Delete all occurrences of data 
     * @param data element to delete 
     * 
     */
    public void deleteAll(int data) {
        Node current = head;
        if (head == null) { // list is empty
            return;
        }
        //while loop to delete consecutive occurances of data
        while(head.data == data) { // if we want to delete the head , make next node head
            head = head.next;
            size -= 1;
        }
        while(current.next != null) {
            //while loop to delete consecutive occurances of data
            while (current.next.data == data) { 
                current.next = current.next.next;
                size -= 1;
            }
            current = current.next;
        }
    }

    public void reverse() {

    }



    /**
     * Prints the whole linked list 
     * Time Complexity : O(n)
     */
    public void print() {

        if(head == null) { //list is empty
            return;
        }
        Node current = head;
        while (current.next != null) {
            System.out.print(current.data + "->");
            current = current.next;
        }
        System.out.print(current.data + "\n"); 
    }
}

public class LinkedListTest {

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
        LinkedList lt = new LinkedList();
        lt.print();
        lt.add(3);
        lt.add(5);
        lt.add(6);
        lt.print();
        lt.add(4, 1);
        lt.print();
        lt.add(4, 7);// 7 is an invalid index
        lt.add(8, 3);
        lt.add(8, 4);
        lt.print();
        System.out.println("Position : " + lt.find(8));
        lt.delete(5);
        lt.print();
        lt.deleteAll(8);
        lt.print();
        System.out.println("Size : " + lt.getSize());
    }

}

Answer 1

您实施的时间复杂度为O(n)，而不是您认为的O(n + n^2)。 虽然嵌套循环是O(n^2)的常见符号，但情况并非总是如此。 重点是迭代次数。 在你的情况下， 如果你在嵌套循环中执行k步骤， 通过k减少外循环中的剩余步骤。 总体， 你仍然会采取n步骤来达到目的。

但是你的实现有一些bug， 并且还可以改进：

• 错误：您指定了current = head，但如果head.data == data则会被删除，而current仍会指向它
• 不需要嵌套循环。对头部进行特殊处理后，您只需按照节点并删除匹配的项目
即可

像这样：

public void deleteAll(int data) {
    while (head != null && head.data == data) {
        head = head.next;
        size -= 1;
    }

    if (head == null) {
        return;
    }

    Node current = head;
    while (current.next != null) {
        if (current.next.data == data) {
            current.next = current.next.next;
            size -= 1;
        } else {
            current = current.next;
        }
    }
}

顺便说一句，头部的特殊处理可能有点烦人。 一个优雅的选择是创建一个指向head的虚拟：

public void deleteAll(int data) {
    Node dummy = new Node();
    dummy.next = head;

    Node current = dummy;
    while (current.next != null) {
        if (current.next.data == data) {
            current.next = current.next.next;
            size -= 1;
        } else {
            current = current.next;
        }
    }

    head = dummy.next;
}

Answer 2

请注意，除了那些原始类型之外，Java中的所有内容都是引用。因此current仍保留原始head。

没有必要将连续出现作为特殊情况处理。通过链接列表进行简单的迭代和判断O(n) - 线性复杂度将完全符合您的要求。

此外，嵌套循环并不一定意味着它肯定会花费O（n ^ 2）。如果您每次都移动current，它仍然会在线性时间内遍历链接列表。

另一个个人建议：如果在处理链表时需要写node.next.next之类的内容，则需要设置另一个引用或重构代码。