【9】138. Copy List with Random Pointer

138. Copy List with Random Pointer 

• Total Accepted: 95300
• Total Submissions: 357941
• 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.

Solution: 

Map the old list node to the new node.

/**
 * 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*> old2new;
        RandomListNode* tmp = head;
        RandomListNode* dummy = new RandomListNode(-1);//从dummy开始，
        RandomListNode* curr = dummy;//很容易建立next指针。curr的next指针只要指向current node即可
        while(tmp){
            RandomListNode* newnode = new RandomListNode(tmp -> label);
            //newnode -> next
            old2new[tmp] = newnode;
            curr -> next = newnode;
            curr = curr -> next;//curr也要往前走，才能完成整个链的链接
            tmp = tmp -> next;
        }
        tmp = head;
        while(tmp){
            //old2new[tmp] -> next = old2new[tmp -> next];
            if(tmp -> random){//要判断current node有无random指针
                old2new[tmp] -> random = old2new[tmp -> random];
            }
            tmp = tmp -> next;
        }
        return dummy -> next;//dummy不能变，否则return的就不是一整个链了
    }
};