题目:
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)
答案:
解法一:
看了别人的思路写的,思路如下:(来源见水印)
代码如下:
/*
public class RandomListNode {
int label;
RandomListNode next = null;
RandomListNode random = null;
RandomListNode(int label) {
this.label = label;
}
}
*/
public class Solution {
public RandomListNode Clone(RandomListNode pHead)
{
if(pHead==null){
return null;
}
//next
RandomListNode current = pHead;
while(current!=null){
RandomListNode node = new RandomListNode(current.label);
node.next = current.next;
current.next = node;
current = node.next;
}
//random
current = pHead;
while(current!=null){
current.next.random = (current.random==null?null:current.random.next);
current = current.next.next;
}
//result
current = pHead;
RandomListNode result = pHead.next;
while(current!=null){
RandomListNode cloneNode = current.next;
current.next = cloneNode.next;
cloneNode.next = cloneNode.next==null?null:cloneNode.next.next;
current = current.next;
}
return result;
}
}