【发布时间】:2011-06-26 10:12:50
【问题描述】:
我正在研究数据结构和链表,但我不了解如何制作链表的副本。有人可以解释一下吗,可能使用伪代码或 C 代码?
【问题讨论】:
标签: c algorithm data-structures linked-list
我正在研究数据结构和链表,但我不了解如何制作链表的副本。有人可以解释一下吗,可能使用伪代码或 C 代码?
【问题讨论】:
标签: c algorithm data-structures linked-list
复制链表的逻辑是递归的,基于以下观察:
如果你用 C++ 对链表进行编码,这会非常干净:
struct Node {
int value;
Node* next;
};
Node* Clone(Node* list) {
if (list == NULL) return NULL;
Node* result = new Node;
result->value = list->value;
result->next = Clone(list->next);
return result;
}
【讨论】:
您了解如何将新节点添加到现有列表中吗?您是否了解如何遍历(即迭代)列表?复制列表只是同时执行这两个操作(遍历 ListA;对于每个元素,复制该元素并将其作为新节点添加到 ListB)。
【讨论】:
这个帖子的答案已经给出并被接受——一切都很好! 但是,如果有人正在寻找 C# 中的迭代方法,这里是:
Node 类:
public class Node
{
public Node(int val)
{
Val = val;
}
public Node Next { get; set; }
public int Val { get; }
}
这里是迭代实现:
public Node CopyLinkedListIteratively(Node head)
{
// validation:
if (head == null) return null;
// current node is always a 'new' node with value.
Node currentNode = new Node(head.Val);
// set copyList and previous to current node to start with - which is true at this point in time!
Node copyList = currentNode;
Node previous = currentNode;
// move head one step forward as we already have copied its value.
head = head.Next;
// keep moving until we hit a null reference which is the end.
while (head != null)
{
currentNode = new Node(head.Val); // create a new node every time as we move forward.
previous.Next = currentNode; // set previous node's next to current node as previous node itself is one step behind the current.
previous = previous.Next; // move prev pointer forward
head = head.Next; // move head pointer forward as well
}
// return the reference to copyList.
// copyList and previous both started off pointing to the currentNode, then in the while loop
// previous kept moving forward till all nodes are copied.
// copyList reference never moved from its position so its still pointing at the start.
return copyList;
}
时间复杂度:O(n)
空间复杂度:O(n)
其中 n = 链表中的节点数。
个人偏好使用递归或迭代方法,但是我建议在使用递归时考虑函数调用堆栈。
单元测试:
[Test]
public void CopyLinkedListIterativelyTest()
{
Node head = new Node(1);
head.Next = new Node(2);
head.Next.Next = new Node(3);
head.Next.Next.Next = new Node(4);
head.Next.Next.Next.Next = new Node(5);
var actual = runner.CopyLinkedListIteratively(head);
while (actual != null)
{
Assert.AreEqual(head.Val, actual.Val);
actual = actual.Next;
head = head.Next;
}
}
【讨论】: