剑指offer-复杂链表的复制

输入一个复杂链表（每个节点中有节点值，以及两个指针，一个指向下一个节点，另一个特殊指针指向任意一个节点），返回结果为复制后复杂链表的head。

方法1：先按next建立好初始链表，然后从头结点开始遍历原链表每个节点的random指针需要走几步，在新建立的链表中走同样的步数建立random指针。

/*
struct RandomListNode {
    int label;
    struct RandomListNode *next, *random;
    RandomListNode(int x) :
            label(x), next(NULL), random(NULL) {
    }
};
*/
class Solution {
    int find(RandomListNode* a,RandomListNode* b){
        int sum=0;
        while(a!=b){
            a=a->next;
            sum++;
        }
        return sum;
    }
public:
    RandomListNode* Clone(RandomListNode* pHead)
    {
        if(pHead==NULL)
        return NULL;
         RandomListNode* he=pHead;
        RandomListNode* head1=new RandomListNode(he->label);
        RandomListNode* p=head1;
        RandomListNode* q=pHead;
        while(q->next){
            p->next=new RandomListNode(q->next->label);
            p=p->next;
            q=q->next;
        }
        p=head1;
        q=pHead;
        while(p){
            if(q->random==NULL){
                p->random=NULL;
            }
            else{
              RandomListNode* po=head1;
            for(int i=0;i<find(pHead,q->random);i++){
                po=po->next;
            }
            p->random=po;
            }
            p=p->next;
            q=q->next;
        }
        return head1;
    }
};

方法2：

1、复制每个节点，如：复制节点A得到A1，将A1插入节点A后面

2、遍历链表，A1->random = A->random->next;

3、将链表拆分成原链表和复制后的链表




 RandomListNode* Clone(RandomListNode* pHead)
    {
        if(!pHead) return NULL;
        RandomListNode *currNode = pHead;
        while(currNode){
            RandomListNode *node = new RandomListNode(currNode->label);
            node->next = currNode->next;
            currNode->next = node;
            currNode = node->next;
        }
        currNode = pHead;
        while(currNode){
            RandomListNode *node = currNode->next;
            if(currNode->random){               
                node->random = currNode->random->next;
            }
            currNode = node->next;
        }
        //拆分
        RandomListNode *pCloneHead = pHead->next;
        RandomListNode *tmp;
        currNode = pHead;
        while(currNode->next){
            tmp = currNode->next;
            currNode->next =tmp->next;
            currNode = tmp;
        }
        return pCloneHead;
    }
};

方法3：

map关联：将一对复制的节点<N,N'>关联起来，可以直接根据N->random找到N'->random

class Solution {
public:
    RandomListNode* Clone(RandomListNode* pHead)
    {
        if(pHead==NULL) return NULL;
 
        map<RandomListNode*,RandomListNode*> m;
        RandomListNode* pHead1 = pHead;
        RandomListNode* pHead2 = new RandomListNode(pHead1->label);
        RandomListNode* newHead = pHead2;
        m[pHead1] = pHead2;
        while(pHead1){
            if(pHead1->next) pHead2->next = new RandomListNode(pHead1->next->label);
            else pHead2->next = NULL;
            pHead1 = pHead1->next;
            pHead2 = pHead2->next;
            m[pHead1] = pHead2;
        }
 
        pHead1 = pHead;
        pHead2 = newHead;
        while(pHead1){
            pHead2->random = m[pHead1->random];
            pHead1 = pHead1->next;
            pHead2 = pHead2->next;
        }
        return newHead;
    }
};