【编程题目】复杂链表的复制☆

76.复杂链表的复制（链表、算法）
题目：有一个复杂链表，其结点除了有一个 m_pNext 指针指向下一个结点外，
还有一个 m_pSibling 指向链表中的任一结点或者 NULL。其结点的 C++定义如下：
struct ComplexNode
{
int m_nValue;
ComplexNode* m_pNext;
ComplexNode* m_pSibling;
};
请完成函数 ComplexNode* Clone(ComplexNode* pHead)，以复制一个复杂链表。

思路：先按照m_pNext复制，再依次遍历链表找m_pSibling

#include <stdio.h>
#include <stdlib.h>typedef struct ComplexNode
{ int m_nValue;ComplexNode* m_pNext;ComplexNode* m_pSibling;
}ComplexNode;ComplexNode*  Clone(ComplexNode* pHead) //完全复制
{if (pHead == NULL){return NULL;}//先复制头结点ComplexNode* chead = (ComplexNode*)malloc(sizeof(ComplexNode));chead->m_nValue = pHead->m_nValue;chead->m_pNext = NULL;chead->m_pSibling = NULL;ComplexNode* x = pHead->m_pNext;ComplexNode* y = chead;while(x != NULL) //按照next信息先创建所有的点并连接
    {y->m_pNext = (ComplexNode*)malloc(sizeof(ComplexNode));y->m_pNext->m_nValue = x->m_nValue;y->m_pNext->m_pNext = NULL;y->m_pNext->m_pSibling = NULL;x = x->m_pNext;y = y->m_pNext;}//复制m_pSibling的指向x = pHead;y = chead;while (x != NULL){if (x->m_pSibling == NULL)  //空指针直接复制
        {y->m_pSibling = NULL;}else //非空指针 从头查找指向第几个节点
        {if (pHead == x->m_pSibling){y->m_pSibling = chead;}else{ComplexNode * xx = pHead;ComplexNode * yy = chead;while (xx->m_pNext != x->m_pSibling){xx = xx->m_pNext;yy = yy->m_pNext;}y->m_pSibling = yy->m_pNext;}    }x = x->m_pNext;y = y->m_pNext;}return chead;
}void createList(ComplexNode* &pHead) //创建并添加next间关系
{int data;scanf("%d", &data);if (data != 0){pHead = (ComplexNode*)malloc(sizeof(ComplexNode));pHead->m_nValue = data;pHead->m_pNext = NULL;pHead->m_pSibling = NULL;createList(pHead->m_pNext);}
}void addsibling(ComplexNode* pHead) //添加pSibling间关系
{int len = 0;ComplexNode* x = pHead;while (x != NULL){len++;x = x->m_pNext;}x = pHead;while (x != NULL){int n = 0; //数字表示p_Sibling指向链表中的第n个结点printf("please input a number between 1 - %d or 0(means NULL)", len);scanf("%d", &n);if (n == 0){x->m_pSibling = NULL;}else{ComplexNode* y = pHead;n = n - 1;while (n != 0){y = y->m_pNext;n--;}x->m_pSibling = y;}x = x->m_pNext;}}int main()
{ComplexNode* pHead = NULL;createList(pHead);addsibling(pHead);ComplexNode* cHead = Clone(pHead);return 0;
}

网上看答案发现自己的方法很复杂，有一种很巧妙的方法：

http://blog.csdn.net/zhaojinjia/article/details/9313275

方法三：

比较巧妙，只需遍历3次链表，时间代价为O(n)，空间代价为0，分3步

1：遍历一遍原始链表，复制结点N对应的N'，将其插入到结点N的后面，如下图所示

2：确定每个m_pSibling指针的指向，只需遍历一遍链表即可确定每个结点的m_pSibling指针的指向，得到如下图结构

3：再次遍历一遍，将原始链表和复制链表分开，奇数为原始链表，偶数为复制链表，得到如下图型