力扣：随即指针138. 复制带随机指针的链表

复制带随机指针的链表

OJ链接

分析：
该题的大致题意就是有一个带随机指针的链表，复制这个链表但是不能指向原链表的节点，所以每一个节点都要复制一遍
大神思路：
ps:我是学来的

在这里插入图片描述
上代码：

struct Node* copyRandomList(struct Node* head)
{//1.在原链表每个节点的后面复制一个节点struct Node* cur = head;while(cur){//插入struct Node*copy = (struct Node*)malloc(sizeof(struct Node));if(copy == NULL){perror("malloc\n");return NULL;}copy->val = cur->val;struct Node* next = cur->next;cur->next = copy;copy->next = next;//迭代cur = next;} //2.处理randomcur = head;while(cur){struct Node*copy = cur->next;if(cur->random == NULL){copy->random = NULL;}else{copy->random = cur->random->next;//这个思路的点睛之笔}cur = copy->next;//迭代}//3.恢复原链表，链接新链表  删除＋尾插 cur=head;struct Node* copyhead = NULL;struct Node* copytail = NULL;while(cur){struct Node* copy = cur->next;struct Node* next = copy->next;//用来还原原链表//尾插：链接新链表//空链表（第一次尾插）if(copyhead == NULL){copyhead = copytail = copy;}else{copytail->next = copy;//尾插copytail = copytail->next;//迭代 }//删除：恢复原链表//free(cur->next);//此处不用freecur->next = next;cur = cur->next;//迭代}return copyhead;
}