Question

我需要理解简单的例子

 LinkedList list = new LinkedList();
        list.add("J");
        list.add("A");
        list.add("V");
        list.add("A");

我有简单的LinkedList，需要将它传递给方法反向

public void reverse(LinkedList list) {

}

将返回反向新列表

我不需要任何ArraysUtils来反转它

反转输出的简短和实践方法是什么？

同样需要理解简单数组。

Answer 1

您可以使用Collections class。

Collections.reverse(list);

在此操作后，您的列表将被撤消。

Answer 2

试试这个。

public LinkedList<String> reverse(LinkedList list) {
LinkedList reverseList=null;

for(int i=list.size();i>0;i--)
{
reverseList.add(list.get(i))
}
return reverseList;
}

Answer 3

集合类的反向方法是我猜你需要的。您可以根据需要更改实施

/**
 * Reverses the order of the elements in the specified list.<p>
 *
 * This method runs in linear time.
 *
 * @param  list the list whose elements are to be reversed.
 * @throws UnsupportedOperationException if the specified list or
 *         its list-iterator does not support the <tt>set</tt> operation.
 */
public static void reverse(List<?> list) {
    int size = list.size();
    if (size < REVERSE_THRESHOLD || list instanceof RandomAccess) {
        for (int i=0, mid=size>>1, j=size-1; i<mid; i++, j--)
            swap(list, i, j);
    } else {
        ListIterator fwd = list.listIterator();
        ListIterator rev = list.listIterator(size);
        for (int i=0, mid=list.size()>>1; i<mid; i++) {
    Object tmp = fwd.next();
            fwd.set(rev.previous());
            rev.set(tmp);
        }
    }
}

Answer 4

public void reverse(LinkedList list) {
    //run till middle and swap start element with end element and so on
    for(int i = 0, mid = list.size()/2, j = list.size() - 1; i < mid; i++, j--)
        list.set(i, list.set(j, list.get(i)));//swap
}

注意：List.set(..,..)方法先前返回指定位置的元素

Answer 5

public ListNode reverseList(ListNode head) {
    if(head == null) return head;

    Stack<ListNode> stack = new Stack<ListNode>();
    while(head != null) {
        stack.push(head);
        head = head.next;
    }

    ListNode dummy = new ListNode(0);
    head = dummy;
    while(!stack.isEmpty()) {
        ListNode current = stack.pop();
        head.next = new ListNode(current.val);
        head = head.next;
    }

    return dummy.next;
}

java中的反向链接列表

5 个答案: