数据结构（7）单链表算法题OVA

随机链表的复制

1、题目描述

https://leetcode.cn/problems/copy-list-with-random-pointer

2、思路分析 

第一步、在原链表的基础上拷贝节点

第二步、置random指针

 只要节点不为空，那么就满足以下这个结论：copy -> random = cur -> random -> next

第三步、断开新旧链表 

定义指针pcur指向原链表的头结点，用来遍历整个链表，再定义拷贝链表的头尾指针copyHead和copyTail。 

3、参考代码 

/**
 * Definition for a Node.
 * struct Node {
 *     int val;
 *     struct Node *next;
 *     struct Node *random;
 * };
 */
typedef struct Node Node;
Node* buyNode(int x)
{
    Node* newNode = (Node*)malloc(sizeof(Node));
    newNode->val = x;
    newNode->next = newNode->random = NULL;

    return newNode;
}
void AddNode(Node* head)
{
    Node* pcur = head;
    while(pcur)
    {
        Node* newnode = buyNode(pcur->val);
        Node* next = pcur->next;
        newnode->next = next;
        pcur->next = newnode;
        pcur = next;
    }
}
void setRandom(Node* head)
{
    Node* pcur = head;
    while(pcur)
    {
        Node* copy = pcur->next;
        if(pcur->random)
        {
            copy->random = pcur->random->next;
        }
        pcur = copy->next;
    }
}
struct Node* copyRandomList(struct Node* head) 
{
    if(head == NULL)
    {
        return head;
    }
	//在原链表的基础上拷贝节点并插入到原链表中
    AddNode(head);
    //设置random指针
    setRandom(head);
    //断开新链表
    Node* pcur = head;
    Node* copyHead, *copyTail;
    copyHead = copyTail = pcur->next;
    while(copyTail->next)
    {
        pcur = copyTail->next;
        copyTail->next = pcur->next; 
        copyTail = copyTail->next;
    }
    return copyHead;
}