【LeetCode】138. Copy List with Random Pointer

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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.

题解：复制一个随机链表，本题复制时先用map标记原地址到复制地址的映射，定义一个函数来判断是否在map以及new操作，之后直接复制next和random同时进行切勿分开

代码：

/**
 * 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 *getClone(RandomListNode *head)
    {
        if(head==NULL) return NULL;
        if(mp.find(head)!=mp.end()) return mp[head];
        RandomListNode *p=new RandomListNode(head->label);
        mp[head]=p;
        return p;
    }
    RandomListNode *copyRandomList(RandomListNode *head){
        
        if(head==NULL) return NULL;
        RandomListNode *p,*q=head;
        p=new RandomListNode(q->label);
        mp[q]=p;
        while(q!=NULL){
            
            p->next=getClone(q->next);
            p->random=getClone(q->random);
            mp[q]=p;
            p=p->next;
            q=q->next;
        }
        return mp[head];
    }
private:
    map<RandomListNode*,RandomListNode*> mp;
    
};

递归方法：

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