程序员面试题精选100题(49)-复杂链表的复制

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;
};
下图是一个含有5个结点的该类型复杂链表。图中实线箭头表示m_pNext指针，虚线箭头表示m_pSibling指针。为简单起见，指向NULL的指针没有画出。



首先在原链表的每个节点后都创建一个对应的节点，初始化的时候m_pSibling=NULL

void CloneNodes(ComplexNode* pHead)
{
    ComplexNode* pNode = pHead;
    while(pNode != NULL)
    {
        ComplexNode* pCloned = new ComplexNode();
        pCloned->m_nValue = pNode->m_nValue;
        pCloned->m_pNext = pNode->m_pNext;
        pCloned->m_pSibling = NULL;
        pNode->m_pNext = pCloned;
        pNode = pCloned->m_pNext;
    }
}
结果如下：


开始设置新插入的节点的m_pSibling的值，A中新的节点1的m_pSibling 等于原节点中1的m_pSibling的m_pNext

void ConnectSiblingNodes(ComplexNode* pHead)
{
    ComplexNode* pNode = pHead;
    while(pNode != NULL)
    {
        ComplexNode* pCloned = pNode->m_pNext;
        if(pNode->m_pSibling != NULL)
        {
            pCloned->m_pSibling = pNode->m_pSibling->m_pNext;
        }
        pNode = pCloned->m_pNext;
    }
}

构造完新链表之后，提取链表

ComplexNode* ReconnectNodes(ComplexNode* pHead)
{
    ComplexNode* pNode = pHead;
    ComplexNode* pClonedHead = NULL;
    ComplexNode* pClonedNode = NULL;
    if(pNode != NULL)
    {
        pClonedHead = pClonedNode = pNode->m_pNext;
        pNode->m_pNext = pClonedNode->m_pNext;
        pNode = pNode->m_pNext;
    }
    while(pNode != NULL)
   {
        pClonedNode->m_pNext = pNode->m_pNext;
        pClonedNode = pClonedNode->m_pNext;
        pNode->m_pNext = pClonedNode->m_pNext;
        pNode = pNode->m_pNext;
    }
    return pClonedHead;
}