138. Copy List with Random Pointer

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• Total Submissions: 368319
• Difficulty: Medium
• Contributors: Admin

A linked list is given such that each node contains an additional random pointer which could point to any node in the list or null.

Return a deep copy of the list.

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来源： https://leetcode.com/problems/copy-list-with-random-pointer/?tab=Description

分析

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本题实际上就是有向图的深拷贝。

方法一

时间复杂度O(3N)

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/**  * Definition for singly-linked list with a random pointer.  * struct RandomRandomListNode {  *     int label;  *     RandomRandomListNode *next, *random;  *     RandomRandomListNode(int x) : label(x), next(NULL), random(NULL) {}  * };  */ class Solution { public:     RandomListNode *copyRandomList(RandomListNode *head) {         if(head == NULL) return NULL;                   RandomListNode* p = head;         //copy every node and link to one's next position         while(p != NULL){             RandomListNode* next = p->next;             RandomListNode* newNode = new RandomListNode(p->label);             p->next = newNode;             newNode->next = next;             p = next;         }                   //every new node's random link to orignal link's next node         p = head;         while(p != NULL){            if(p->random != NULL)p->next->random = p->random->next;            p = p->next->next;         }                   //split the list         RandomListNode dummy1(0), dummy2(0);         RandomListNode* p1 = &dummy1, * p2 = &dummy2;         p = head;         while(p != NULL){             p1->next = p;              p1 = p1->next;             p2->next = p->next;              p2 = p2->next;             p = p->next->next;             if(p == NULL){                 p1->next = NULL;                 p2->next = NULL;             }         }         head = dummy1.next;         return dummy2.next;     } };

方法二

使用map去做，建立 新旧节点的 key-value 对，然后拷贝连接关系即可

（用C++写会runtime error， Java就没问题，不懂）

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/**  * Definition for singly-linked list with a random pointer.  * struct RandomListNode {  *     int label;  *     RandomListNode *next, *random;  *     RandomListNode(int x) : label(x), next(NULL), random(NULL) {}  * };  */ class Solution { public:     RandomListNode *copyRandomList(RandomListNode *head) {         if(head == NULL) return NULL;                   unordered_map<RandomListNode*, RandomListNode*> map;         RandomListNode* p = head;         while(p != NULL){             map.insert({p,new RandomListNode(p->label)});             p = p->next;         }                   for(auto p: map){             RandomListNode *newNode = (p.second);             newNode->next = map[p.first->next];             newNode->random = map[p.first->random];         }         return map[head];     } };

方法三

使用迭代，原理和方法二一样

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class Solution { public:     unordered_map<RandomListNode*, RandomListNode*>mp;     RandomListNode *copyRandomList(RandomListNode *head)      {         if(!head) return NULL;         if(mp[head]!=NULL) return mp[head];         mp[head] = new RandomListNode(head->label);         mp[head] -> next = copyRandomList(head->next);         mp[head] -> random = copyRandomList(head->random);         return mp[head];     } };

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