1,问题的描述
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)
2,解题思路(转自牛客网)
(1)遍历链表,复制每个节点并将该复制后的新节点放至旧节点之后,先不用管每个节点的随机节点
(2)重新遍历链表,复制旧节点的随机指针给新节点
(3)拆分链表,将链表拆分为原链表和复制后的链表
3,源码
/*
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;
//1,遍历链表,复制每个节点并将该复制后的新节点放至旧节点之后
RandomListNode currentHead = pHead;
while(currentHead!=null) {
RandomListNode copyNode = new RandomListNode(currentHead.label);
RandomListNode nextNode = currentHead.next;
copyNode.next = nextNode;
currentHead.next = copyNode;
currentHead = nextNode;
}
//2,重新遍历链表,复制旧节点的随机指针给新节点
currentHead = pHead;
while(currentHead!=null) {
currentHead.next.random = currentHead.random==null?null:currentHead.random.next;
currentHead = currentHead.next.next;
}
//3,拆分链表,将链表拆分为原链表和复制后的链表
currentHead = pHead;
RandomListNode cloneHead = currentHead.next;
while(currentHead!=null) {
RandomListNode cloneNode = currentHead.next;
currentHead.next = cloneNode.next;
cloneNode.next = currentHead.next==null?null:currentHead.next.next;
currentHead = currentHead.next;
}
return cloneHead;
}
}