剑指offer系列19：复杂链表的复制

这道题是典型的分治法，将一个大问题分解成几小步解决。一定要注意在使用指针的时候指针指向是否为空的问题。在指针指向时，可以将一个指向为空的指针作为赋值来写，但是空指针不能指向任何地方（指向空也不行），这点一定要记住。

#include<iostream>
#include<vector>
using namespace std;
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 == NULL)
            return NULL;
        //3步
        //第一步，将复杂链表复制成double
        nodeclone(pHead);
        connect(pHead);
        return reconnect(pHead);
    }
    void nodeclone(RandomListNode* Head)
    {
        RandomListNode*p = Head;
        while (p != NULL)
        {
            //插入节点
            //cur = p->next;
            RandomListNode *curnode = new RandomListNode();
            curnode->label = p->label;
            curnode->next = p->next;
            curnode->random = NULL;//这一句必须要写？
            p->next = curnode;
            p = curnode->next;//悬浮指针？
        }
    }
    void connect(RandomListNode* Head)
    {
        //第二步：将复制出的结点的随机指针指向该有的位置
        RandomListNode*p = Head;
        while (p != NULL)//在指针访问的时候，时刻注意是否为空
        {
            RandomListNode *curnode = p->next;

            if (p->random != NULL)
            {
                curnode->random = p->random->next;
            }

            p = curnode->next;

        }
    }
    RandomListNode *reconnect(RandomListNode* Head)
    {
        //第三步：将链表拆为两部分
        RandomListNode*p = Head;
        RandomListNode *copy = p->next;
        while (p!= NULL)
        {
            RandomListNode *curnode = p->next;
            if (curnode != NULL)
            {
                p->next = curnode->next;
            }

            p = curnode;
        }
        return copy;
    }
};
int main()
{
    RandomListNode list[5];
    list[0].label = 1;
    list[0].next = &list[1];
    list[0].random = &list[2];

    list[1].label = 2;
    list[1].next = &list[2];
    list[1].random = &list[4];

    list[2].label = 3;
    list[2].next = &list[3];
    list[2].random = NULL;

    list[3].label = 4;
    list[3].next = &list[4];
    list[3].random = &list[1];

    list[4].label = 5;
    list[4].next =NULL;
    list[4].random = NULL;

    Solution solu;
    RandomListNode *re = solu.Clone(list);
    int count=0;
    while (re != NULL)
    {
        //cout << re->label << " "<<re->random->label<<",";
        cout << re->label<<" ";
        if (re->random != NULL)
            cout << re->random->label;
        cout << endl;
        count++;
        re = re->next;
    }
    //cout << endl;
    cout << "number of array:"<<count << endl;
    return 0;
}