给定一个双向循环链表和一个关键字，我们需要在链表中搜索关键字，并在找到时给出适当的消息。假设我们有一个具有特定字符的链表，并且我们需要在其中搜索一个元素。所以让我们从下面的链表开始 -

<-> 5 <-> 8 <-> 9 <-> 2 <-> 4 <->

我们将使用4作为寻找给定问题解决方案的关键。双向链表没有固定的头部，所以我们将从任意节点开始，然后将该节点标记为头部，直到我们再次遇到这个头部，我们对链表进行线性搜索，并搜索关键字。

让我们看一些输入输出的场景 -

假设我们有一个双向循环链表，其中有5个节点 <-> 3 <-> 4<-> 5<-> 6<-> 7<->，要查找的元素是6。

Input = <-> 3 <-> 4<-> 5<-> 6<-> 7<-> key=6
Output = Element found

让我们考虑另一种情况，即在双向循环链表中没有要搜索的元素。

Input = <-> 10<->20<->30<->40<->50<-> key=100
Output = Element not found

算法

以下是接近的步骤。

• 实现一个链表，并通过在链表的每个节点中分配前向节点来传递值。

• 将节点的前一部分分配给最后一个节点的下一部分。

• 将节点的每个前一部分分配给节点的下一部分。

• 将关键元素传递给检查它是否存在于双向循环链表中的关键元素。

• 如果键存在于双向循环链表中，则返回true。

• Else, it returns false.

Example

示例

以下是在双向链表中执行搜索操作的C++实现代码：

#</span>include</span> <iostream></span></span>
#</span>include</span> <vector></span></span>
using</span> namespace</span> std;</span>
class</span> Node</span> {</span>
   public</span>:</span>
   int</span> val;</span>
   Node *</span>left,</span> *</span>right;</span>
   Node</span>(</span>int</span> val)</span> {</span>
      this</span>-></span>val =</span> val;</span>
   }</span>
}</span>;</span>
bool</span> solve</span>(</span>Node*</span> root,</span> int</span> key)</span> {</span>
   Node*</span> copy =</span> root;</span>
   do</span> {</span>
      if</span>(</span>copy-></span>val ==</span> key)</span> return</span> true</span>;</span>
      copy =</span> copy-></span>right;</span>
   }</span>while</span>(</span>copy!=</span>root)</span>;</span>
   return</span> false</span>;</span>
}</span>
int</span> main</span>(</span>)</span> {</span>
   // assigning the forward node in each node of the linked list</span>
   Node*</span> phead =</span> new</span> Node</span>(</span>5</span>)</span>;</span>
   phead-></span>right =</span> new</span> Node</span>(</span>8</span>)</span>;</span>
   phead-></span>right-></span>right =</span> new</span> Node</span>(</span>9</span>)</span>;</span>
   phead-></span>right-></span>right-></span>right =</span> new</span> Node</span>(</span>2</span>)</span>;</span>
   phead-></span>right-></span>right-></span>right-></span>right =</span> new</span> Node</span>(</span>4</span>)</span>;</span>
   phead-></span>right-></span>right-></span>right-></span>right-></span>right =</span> phead;</span>
 
   // assignment of the previous node in each node in the linked list</span>
 
   // assigning the previous of the head to the last element</span>
   phead-></span>left =</span> phead-></span>right-></span>right-></span>right-></span>right;</span>

   // assigning the left node in each node of the linked list</span>
   phead-></span>right-></span>left =</span> phead;</span>
   phead-></span>right-></span>right-></span>left =</span> phead-></span>right;</span>
   phead-></span>right-></span>right-></span>right-></span>left =</span> phead-></span>right-></span>right;</span>
   phead-></span>right-></span>right-></span>right-></span>right-></span>left =</span> phead-></span>right-></span>right-></span>right;</span>
   if</span>(</span>solve</span>(</span>phead,</span> 4</span>)</span>)</span> cout <<</span> "Element present"</span>;</span> else</span> cout <<</span> "Element not present"</span>;</span>
   return</span> 0</span>;</span>
}</span>

Output

Element present

Explanation

解释

关键字4存在于双向链表中。

结论

在一个双向循环链表中，我们可以从任何位置开始，因为没有固定的头部和尾部。在上述方法中，我们有一个“头部”，它是一个伪头部，我们从这里开始搜索。上述算法的时间复杂度是O(n)，因为是线性搜索。