题解 | #其实不难，难就难在不能用题目的链表节点直接赋值给新开辟的链表 复杂链表的复制#

/*
struct RandomListNode {
    int label;
    struct RandomListNode *next, *random;
    RandomListNode(int x) :
            label(x), next(NULL), random(NULL) {
    }
};
*/
class Solution {
public:
    RandomListNode* Clone(RandomListNode* pHead) {
        RandomListNode* p1 = pHead;vector<int> values;
         while(p1){
            values.push_back(p1->label);
            p1 = p1->next;
        }
      /*根据数组重建新的单链表*/
        RandomListNode* head = new RandomListNode(NULL);
        for(int i =values.size()-1;i>=0;--i){
           RandomListNode* p2 = new RandomListNode(values[i]);
           p2->next = head->next;
           head->next = p2;
        }
      /*构建复杂链表*/
        RandomListNode* p3 = head->next;RandomListNode* p4 = pHead;
        while(p3){
            if(p4->random){
               RandomListNode* temp = head->next;
              /*找出题目链表对应的random的节点 */
               while(temp && temp->label != p4->random->label){
                   temp = temp->next;
               }
              /*此时temp 指向 新链表的random,赋值给p3就完了*/
                p3->random = temp;
            }
            p3 = p3->next;
            p4 = p4->next;
        }
        return head->next;
    }
};