【LeetCode & 剑指offer刷题】链表题8：35 复杂链表的复制（138. Copy List with Random Pointer）
【LeetCode & 剑指offer 刷题笔记】目录（持续更新中...）
138. Copy List with Random Pointer
A linked list is given such that each node contains an additional random pointer which could point to any node in the list or null.
Return a deep copy of the list.
/**
* Definition for singly-linked list with a random pointer.
* struct RandomListNode {
* int label;
* RandomListNode *next, *random;
* RandomListNode(int x) : label(x), next(NULL), random(NULL) {}
* };
*/
/*
问题：复制带随机指针的链表
方法：在原链表内部复制链表结点，由内部关系，复制随机指针连接关系，再进行拆分
*/
class Solution
{
public:
RandomListNode *copyRandomList(RandomListNode *head)
{
//第一步：复制结点与next指针
RandomListNode* cur = head;
while(cur != nullptr)
{
RandomListNode* node = new RandomListNode(cur->label); //构造函数会将next和random指针赋值为null
node->next = cur->next;//新结点与旧结点连接
cur->next = node; //旧结点与新结点连接
cur = node->next; //下个旧结点
}
//第二步：复制random指针
cur = head;
while(cur != nullptr)
{
if(cur->random != nullptr) cur->next->random = cur->random->next; //复制random指针 (cur->next为新结点)
cur = cur->next->next; //下个旧结点
}
//第三步：拆分链表
cur = head;
RandomListNode prehead(0); //创建整个链表的空头结点,方便处理
prehead.next = head;
RandomListNode* newcur = &prehead;
while(cur != nullptr)
{
cur->next = cur->next->next;
newcur = newcur->next; //更新指针
cur = cur->next; //cur走在newcur前面，方便判断
}
return prehead.next;
}
};
