我相信我有关于如何删除链表中项目的代码，但是我不确定如何做到这一点，以便将所有出现的值都删除。我需要在哪里进行一些更改以检查列表中的所有值？
我试图做一个备用或假人指向头部，但是我不确定该去哪里。
public class LinkList {
private Link first; // ref to first link on list
// -------------------------------------------------------------
public LinkList() // constructor
{
first = null; // no links on list yet
}
// -------------------------------------------------------------
public void insertFirst(int id, double dd) {
Link newLink = new Link(id, dd);
newLink.next = first; // it points to old first link
first = newLink; // now first points to this
}
// -------------------------------------------------------------
public Link find(int key) // find link with given key
{ // (assumes non-empty list)
Link current = first; // start at 'first'
while (current.iData != key) // while no match,
{
if (current.next == null) // if end of list,
{
return null; // didn't find it
} else // not end of list,
{
current = current.next; // go to next link
}
}
return current; // found it
}
// -------------------------------------------------------------
public void displayList() // display the list
{
System.out.print("List (first-->last): ");
Link current = first; // start at beginning of list
while (current != null) // until end of list,
{
current.displayLink(); // print data
current = current.next; // move to next link
}
System.out.println("");
}
// -------------------------------------------------------------
public Link removeAll(int n) // delete link with given key
{ // (assumes non-empty list)
Link current = first; // search for link
Link previous = first;
while (current.iData != n) {
if (current.next == null) {
return null; // didn't find it
} else {
previous = current; // go to next link
current = current.next;
}
}
if (current == first) // if first link,
{
first = first.next; // change first
} else // otherwise,
{
previous.next = current.next; // bypass it
}
return current;
}
}
我希望删除给定键的所有值，但是我只能删除给定值的一个实例。
答案 0 :(得分：0)
这将删除sess.run中所有出现的for e in range(epochs):
for _ in range(batches):
_, loss_ = sess.run([training_op, loss])
losses.append(loss_)
。如果要通过Link删除，则应该相同。
id == n像这样运行代码：
Link.iData输出：
public Link removeAll(int n)
{
Link head = first;
Link previous = null;
Link current = first;
while (current != null) {
if (current.id == n) {
if (previous == null) {
head = current.next;
} else {
previous.next = current.next;
}
} else {
previous = current; // if we removed current, let previous remain the same
}
current = current.next;
}
first = head;
return head;
}
答案 1 :(得分：0)
我有一个叫ListNode的课程，类似于您的课程。
public class ListNode {
public ListNode next;
public int val;
public ListNode removeAll(int n) {
ListNode newHead = null;
ListNode node = this;
//for keeping track of the node previous to the current node
ListNode prev = null;
//loop through the entire linked list
while (node != null) {
//when you encounter the val == n, delete the node
if (node.val == n) {
if (prev != null){
//this makes the previous node to point the node to the next of the current node
//if 2 -> 1 -> 3 and we have to remove node with key 1 and current node val == 1
// the following code will do this
// 2 -> 3
prev.next = node.next;
}
ListNode next = node.next;
//taking the same example
//this code will break the link : 1->3
node.next = null;
node = next;
} else {
if (newHead == null) {
newHead = node;
}
prev = node;
node = node.next;
}
}
return newHead;
}
}
基本上，您必须遍历整个链表，跟踪当前节点的前一个节点，当您发现一个值/键/数据等于n的节点时，则使前一个节点指向下一个节点当前节点的节点，并断开当前节点到下一个节点的链接。
答案 2 :(得分：0)
我们先删除所有内容。要删除所有出现的内容，您首先需要能够遍历整个列表（以便您可以查找多个匹配项）。遍历整个列表将只是：
public void removeAll(int n) // delete link with given key
{
Link current = first;
Link previous = first;
while (current != null)
{
// simply move to the next Link
previous = current; // store the current node as the previous for the next iteration
current = current.next; // move to the next link
}
}
接下来，让我们添加一个检查以查看当前链接是否为应删除的链接：
public void removeAll(int n) // delete link with given key
{
Link current = first;
Link previous = first;
while (current != null)
{
if (current.iData == n)
{
// To do...delete the current Link
}
else
{
// simply move to the next Link
previous = current; // store the current node as the previous for the next iteration
current = current.next; // move to the next link
}
}
}
一旦找到匹配项，就有两种可能性。链接是第一个链接，或者在列表中更远的位置：
这是更新的代码：
public void removeAll(int n) // delete link with given key
{
Link current = first;
Link previous = first;
while (current != null)
{
if (current.iData == n)
{
if (current == first)
{
current = current.next;
first = current;
previous = current;
}
else
{
current = current.next;
previous.next = current;
}
}
else
{
// simply move to the next Link
previous = current;
current = current.next;
}
}
}
答案 3 :(得分：0)
这是简单的递归实现（此实现保留了初始列表的完整性，并创建了没有指定元素的新列表）：
{{1}}