leetcode_138复制带随机指针的链表

最初版,用原node的next指向新node,从尾部开始更新random,修改了原链表结构未通过

/*
// Definition for a Node.
class Node {
public:
    int val;
    Node* next;
    Node* random;
    
    Node(int _val) {
        val = _val;
        next = NULL;
        random = NULL;
    }
};
*/

class Solution {
public:
    Node* copyRandomList(Node* head) {
        if (head == NULL) {
          return NULL;
        }
        Node *node = new Node(head->val, nullptr, nullptr);
        head->next=node;
        node->next=copyRandomList(temp);
        if(head->random!=NULL){
            node->random=head->random->next;
        }
        return node;
    }
};

2.新链表random完成后修改回去,尾部开始指向就失效了,结果新链表的random会指向错误,未通过

/*
// Definition for a Node.
class Node {
public:
    int val;
    Node* next;
    Node* random;
    
    Node(int _val) {
        val = _val;
        next = NULL;
        random = NULL;
    }
};
*/

class Solution {
public:
    Node* copyRandomList(Node* head) {
        if (head == NULL) {
          return NULL;
        }
        Node *node = new Node(head->val, nullptr, nullptr);
        head->next=node;
        node->next=copyRandomList(temp);
        if(head->random!=NULL){
            node->random=head->random->next;
        }
        head->next=temp;
        return node;
    }
};

新旧链表如果按一对一指针映射必须增加辅助空间做map,参考

https://www.cnblogs.com/reynold-lei/p/3362614.html

 

 将新旧链表二合一,这样就不用多余空间了。

3错版同二的错误,afteconnection必须在全部遍历完成后完成,否则同二一样。

/*
// Definition for a Node.
class Node {
public:
    int val;
    Node* next;
    Node* random;
    
    Node(int _val) {
        val = _val;
        next = NULL;
        random = NULL;
    }
};
*/

class Solution {
public:
    Node* copyRandomList(Node* head) {
        if (head == NULL) {
          return NULL;
        }
        //preConnection
        Node *node = new Node(head->val, head->next, nullptr);
        head->next=node;
        copyRandomList(node->next);
        //randomConnection
        if(head->random!=NULL){
            node->random=head->random->next;
        }
        //afterConnection
        head->next=node->next;
        if(node->next!=NULL){
            node->next=node->next->next;
        }
        return node;
    }
};

4通过版本,在完成一次递归遍历完所有节点后再拆分成两条链

/*
// Definition for a Node.
class Node {
public:
    int val;
    Node* next;
    Node* random;
    
    Node(int _val) {
        val = _val;
        next = NULL;
        random = NULL;
    }
};
*/

class Solution {
public:
    Node* copyRandomList(Node* head) {
        if (head == NULL) {
          return NULL;
        }
        Node* node = pre_copyRandomList(head);
        //after_copyRandomList(head);
        Node* temp;
        while(head->next!=NULL){
            temp=head->next;
            head->next=head->next->next;
            head=temp;
        }
        return node;
    }
    Node* pre_copyRandomList(Node* head) {
        if (head == NULL) {
          return NULL;
        }
        //preConnection
        Node *node = new Node(head->val, head->next, nullptr);
        head->next=node;
        pre_copyRandomList(node->next);
        //randomConnection
        if(head->random!=NULL){
            node->random=head->random->next;
        }
        return node;
    }
    //void after_copyRandomList(Node* head) {
    //    //afterConnection
    //    if(head->next!=NULL){
    //        Node* node=head->next;
    //        head->next=head->next->next;
    //        after_copyRandomList(node);
    //    }
    //}
};

 

posted @ 2021-02-22 17:59  巴啦啦大魔王  阅读(60)  评论(0编辑  收藏  举报