题目描述
输入一个复杂链表(每个节点中有节点值,以及两个指针,一个指向下一个节点,另一个特殊指针指向任意一个节点),返回结果为复制后复杂链表的head。(注意,输出结果中请不要返回参数中的节点引用,否则判题程序会直接返回空)
/*
public class RandomListNode {
int label;
RandomListNode next = null;
RandomListNode random = null;
RandomListNode(int label) {
this.label = label;
}
}
*/
public
class
Solution {
private
void
copyNext(RandomListNode head){
if
(head==
null
){
return
;
}
while
(head!=
null
){
RandomListNode newNode=
new
RandomListNode(head.label);
newNode.random=head.random;
newNode.next=head.next;
head.next=newNode;
head=head.next.next;
}
}
private
void
copyRandom(RandomListNode head){
if
(head==
null
){
return
;
}
while
(head!=
null
){
if
(head.next.random!=
null
){
head.next.random=head.random.next;
}
head=head.next.next;
}
}
private
RandomListNode splitList(RandomListNode head){
if
(head==
null
){
return
null
;
}
RandomListNode newHead=head.next;
while
(head!=
null
){
RandomListNode temp=head.next;
head.next=temp.next;
head=head.next;
if
(temp.next!=
null
){
temp.next=temp.next.next;
}
}
return
newHead;
}
public
RandomListNode Clone(RandomListNode phead)
{
if
(phead==
null
){
return
null
;
}
copyNext(phead);
copyRandom(phead);
return
splitList(phead);
}
}