题解 | #复杂链表的复制# 先复制单向链表，再处理随机指针

/*
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 *tmp = pHead, *res = nullptr;
        RandomListNode *path = nullptr;
        if(tmp != nullptr){
            res = new RandomListNode(tmp->label);
            tmp = tmp -> next;
            path = res;
        }else{
            return nullptr;
        }
        // 完成单向链表的拷贝
        while(tmp != nullptr){
            RandomListNode* tt = new RandomListNode(tmp->label);
            path->next = tt;
            path = path -> next;
            tmp = tmp -> next;
        }
        // 完成random指针的拷贝
        tmp = pHead;
        path = res;
        while(tmp != nullptr){
            if(tmp->random != nullptr){
                path -> random = new RandomListNode(tmp->random->label);
            }
            path = path->next;
            tmp = tmp -> next;
        }
        return res;
    }
};