138. 复制带随机指针的链表

过程：

解着一道题目，就记清楚如下三个步骤：
1、将链表中的每一个结点都进行拷贝，放在节点后面

即：

Node* cur = head;
while(cur)
{
	Node* next = cur->next;
	
	//创建一个结点，并且初始化结点，防止产生野指针
	Node* copy = (Node*)malloc(sizeof(Node));
	copy->next = NULL;
	copy->random = NULL;
	copy->val = cur->val;
	
	//进行链表的重新拼接
	cur->next = copy;
	copy->next = next;
	
	cur = next;
}

2、遍历链表，处理拷贝结点的random

也就是转化为代码：

cur = head;
while(cur)
{
	Node* copy = cur->next;
	if(cur->random)
		copy->random = cur->random->next;
	else
		copy->random = NULL;
	
	cur = cur->next->next;
}

就是让原链表结点random指向的地址，放到新链表中就是让其之后新拷贝的结点指向其random指向结点地址的下一个节点地址。
3、拆解链表，得到复制之后的新链表


转化为代码：

cur = head;
Node* copyhead = head->next;
while(cur)
{
	Node* copy = cur->next;
	Node* next = copy->next;
	
	cur->next = next;
	
	if(next)
		copy->next = next->next;
	else
		copy->next = NULL;

	cur = next;
}

代码实现：

/**
 * Definition for a Node.
 * struct Node {
 *     int val;
 *     struct Node *next;
 *     struct Node *random;
 * };
 */

typedef struct Node Node;
struct Node* copyRandomList(struct Node* head) {
    if(head == NULL)
    {
        return NULL;
    }

    //1、拷贝cur，将拷贝的结点链接到cur与cur->next的之间
	Node* cur = head;
    while(cur)
    {
        Node* copy = (Node*)malloc(sizeof(Node));
        copy->next = NULL;
        copy->random = NULL; 
        copy->val = cur->val;

        Node* next = cur->next;

        copy->next = cur->next;
        cur->next = copy;

        cur = next;
    }

    //2、处理拷贝节点的random
    cur = head;
    while(cur)
    {
        Node* copy = cur->next;
        if(cur->random)
            copy->random = cur->random->next;
        else
            copy->random = NULL;
        
        cur = cur->next->next;
    }

    //3、拆解链表
    cur = head;
    Node* copyHead = head->next;
    while(cur)
    {
        Node* copy = cur->next;
        Node* next = copy->next;

        cur->next = next;
        if(next)
            copy->next = next->next;
        else
            copy->next = NULL;
        
        cur = next;
    }

    return copyHead;
}