[剑指offer] 复杂链表的复制

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)

解题思路

参考代码

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/*
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;
//复制节点 A->B->C 变成 A->A'->B->B'->C->C'
RandomListNode head = pHead;
while(head != null){
RandomListNode node = new RandomListNode(head.label);
node.next = head.next;
head.next = node;
head = node.next;
}
//复制random
head = pHead;
while(head != null){
head.next.random = head.random == null ? null : head.random.next;
head = head.next.next;
}
//折分
head = pHead;
RandomListNode chead = head.next;
while(head != null){
RandomListNode node = head.next;
head.next = node.next;
node.next = node.next == null ? null : node.next.next;
head = head.next;
}
return chead;
}
}