Copy List with Random Pointer

A linked list is given such this each node contains an additional random pointer which could point to all node in the list or null.

Return a deep copy of the list.

Analysis:

The key to this question is actually how to copy the random pointer, find a random pointer point to node, time complexity is O (n). Therefore the total complexity is in O (n^2).

Here is a more ingenious method. Add a node behind each node, and the new node has the same value as the previous node. In this case, the random pointer is easy to copy,

1 /**2 * Definition for singly-linked list with a random pointer.3 * Class Randomlistnode {4 * int label;5 * Randomlistnode Next, random;6 * Randomlistnode (int x) {This.label = x;}7  * };8  */9  Public classSolution {Ten     /** One * @param head:the head of linked list with a random pointer. A * @return: A new head of a deep copy of the list. -      */ -      Publicrandomlistnode copyrandomlist (Randomlistnode listhead) { the         if(Listhead = =NULL)returnListhead; -Randomlistnode current =Listhead; -         //Add Duplicate Nodes -          while(Current! =NULL) { +Randomlistnode node =NewRandomlistnode (Current.label); -Node.next =Current.next; +Current.next =node; ACurrent =Current.next.next; at         } -          -         //set random pointer -Current =Listhead; -          while(Current! =NULL) { -             if(Current.random! =NULL) { inCurrent.next.random =Current.random.next; -             } toCurrent =Current.next.next; +         } -          the         //Restore and Detach *Randomlistnode Newhead =Listhead.next; $Randomlistnode current1 =Listhead;Panax NotoginsengRandomlistnode Current2 =Newhead; -          while(Current1! =NULL) { theCurrent1.next =Current2.next; +Current1 =Current1.next;if (current1! = null) {//Very important, cannot remove it.42 Current2.next = current 1.NEXT;43} -Current2 =Current2.next; $         } $         returnNewhead; -     } -}

Reprint Please specify source: cnblogs.com/beiyeqingteng/

Copy List with Random Pointer