Copy List with Random Pointer(two pointers)
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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.
这道题要返回一个新的,但构造和原来链表相同的一个链表
每个节点有两个指针,一个指向后继节点,还有一个指向某一个节点或者为空,难点在于我们怎样copy随机指针,怎样知道它到底指向哪一个节点
我们可以换一种思路,我们只要知道这个指针到底指向新链表的哪个节点就行,而我们知道随机指针指向原链表的地址,所以我们可以建立一个map,记录两条链表节点一对一的关系
/** * 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==nullptr)return nullptr; RandomListNode prehead(0),*l1=head,*l2=&prehead; map<RandomListNode*,RandomListNode*>m; while(l1) { l2->next=new RandomListNode(l1->label); l2=l2->next; m[l1]=l2; l1=l1->next; } l1=head; l2=prehead.next; while(l1) { if(m.find(l1->random)!=m.end()) l2->random=m[l1->random]; else l2->random=nullptr; l1=l1->next; l2=l2->next; } return prehead.next; }};
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