时间戳

首页 > 解决方案 > 如何通过线性节点中的索引使元素出列

java - 如何通过线性节点中的索引使元素出列

问题描述

如果有人可以提供帮助，我很难找到让该方法删除特定索引上的元素的算法

我试过在下面实现一些，但在逻辑上可能是错误的我也得到了javadoc，这样你就知道算法要求什么

 /**
         * Removes and returns the element that is at place x in the queue.
         * Precondition: x must be less than 5, x must be less than size Note:
         * indexing from 0: 0 == front element, 1 == second element, etc.
         *
         * @param x the passed in index of the element to be removed
         * @return the element removed from the queue
         * @throws EmptyCollectionException if the queue is empty
         * @throws InvalidArgumentException if x > 4, or x > size of collection
         *
         */
        //@Override
    public T dequeue(int x) throws EmptyCollectionException, InvalidArgumentException {
        // exception throw if queue is empty
        if (numNodes == 0) {
            throw new EmptyCollectionException("Empty Collection");
        }
        // exception throw if x > 4 or size
        if (x > 4 || x > size()) {
            throw new InvalidArgumentException("Invalid x");
        }
        T result = null;
        LinearNode<T> temp = front;
        int count;
        while (temp != null) {
            result = temp.getElement();
            if (numNodes == 1) {

                front = null;
                back = null;
                numNodes--;
                return result;
            }
            if (numNodes > 1) {
                if (count == x) {
                    result = result.setNext();
                }
                x++;
                front = temp.getNext();
                front.setPrev(null);
                numNodes--;
                return result;
            }
        }

删除索引 x 的节点

标签： javaalgorithmqueuenodesnodequeue

解决方案

假设您的“LinearNode”类的工作方式类似于 LinkedList 并且具有 getPrevious() 和 getNext() 方法：

public T dequeue(int idx) {
    // Any other bounds checks go here
    LinearNode<T> curNode = front;
    for(int i = 0; i < idx; i++) {
        if(curNode.getNext() != null) {
            curNode = curNode.getNext();
        } else {
            // idx is out of bounds
            throw new InvalidArgumentException("Index is out of bounds");
        }
    }
    LinearNode<T> previousNode = curNode.getPrevious();
    LinearNode<T> nextNode = curNode.getNext();
    if(previousNode != null) {
        previousNode.setNext(nextNode);
    }
    if(nextNode != null) {
        nextNode.setPrevious(previousNode);
    }
    if(curNode.equals(front)) {
        front = nextNode;
    }
    if(curNode.equals(back)) {
        back = previousNode;
    }
    return curNode.getElement();
}

推荐阅读