题解 | #复杂链表的复制#

花了不少时间，首先递归不行，然后看评论方法以为会了，自己写结果发现原来的链表也不能改变，下意识地以为会销毁原来的测试现在的

/*
struct RandomListNode {
    int label;
    struct RandomListNode *next, *random;
    RandomListNode(int x) :
            label(x), next(NULL), random(NULL) {
    }
};
*/
class Solution {
public:
    RandomListNode* Clone(RandomListNode* pHead) {
        if(!pHead) return pHead;
        RandomListNode* cur = pHead;
        while(cur){
            RandomListNode* tmp = new RandomListNode(cur->label);
            tmp->next = cur->next;
            cur->next = tmp;
            cur=tmp->next;
        }
        cur=pHead;
        while(cur){
            if(cur->random)
                cur->next->random=cur->random->next;
            else cur->next->random=cur->random;
            cur=cur->next->next;
        }
        cur=pHead;
        RandomListNode* ans = pHead->next;
        while(cur){
            RandomListNode* t=cur->next->next;
            if(t){
                cur->next->next=t->next;
                cur->next=t;
            }
            else{
                cur->next->next=NULL;
                cur->next=NULL;
            }
            cur=cur->next;
        }
        return ans;
    }
};