随机链表的复制

点击链接做题

思路：

浅拷贝：拷贝值

深拷贝：拷贝空间

1. 在原链表的基础上继续复制链表
2. 置random指针
3. 复制链表和原链表断开

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* phead){
    Node* pcur = phead;
    while(pcur){
        Node* Next = pcur->next;
        //创建新结点，尾插到pcur
        Node* newnode = buyNode(pcur->val);//newnode是复制结点
        pcur->next = newnode;
        newnode->next = Next;

        pcur = Next;
    }
}

struct Node* copyRandomList(struct Node* head) {
    if(head == NULL){
        return NULL;
    }
	//1.原链表上复制结点
    AddNode(head);
    //2.置random
    Node* pcur = head;
    while(pcur){
        Node* copy = pcur->next;
        if(pcur->random != NULL){
            copy->random = pcur->random->next;
        }
        pcur = copy->next;
    }
    //3.断开链表
    //在copy的链表里面设置两个指针：newHead,newTail位于拷贝链表的头
    //pcur往后挪两个到newHead->next，newTail挪到pcur->next
    pcur = head;//让pcur先回到头节点
    Node* newHead, *newTail;
    newHead = newTail = pcur->next;
    while(pcur->next->next){//如果pcur->next->next为空就说明结束了
        pcur = pcur->next->next;
        newTail->next = pcur->next;//让newTail->next指向copy链表的后一个元素
        newTail = newTail->next;//把newTail移到copy链表的后一个元素的位置
    }
    return newHead;//返回新链表newHead到newTail
}